datavorous/entrance-exam-dataset huggingface dataset:
The electric potential $V$ at any point $O\left(x, y, z\right.$ all in metres) in space is given by $V=4 x^2$ volt . The electric field at the point $(1 m, 0,2 m)$ in volt /metre is

Options:
A 8 along negative $X$ - axis
B 8 along positive $X$-axis
C 16 along negative $X$ - axis
D 16 along positive $Z$ - axis

The correct answer is:
8 along negative $X$ - axis

Solution:
The electric potential $V(x, y, z)=4 x^2$ volt
Now $\vec{E}=-\left(\hat{i} \frac{\partial V}{\partial x}+\hat{j} \frac{\partial V}{\partial y}+\hat{k} \frac{\partial V}{\partial z}\right)$
Now $\frac{\partial V}{\partial x}=8 x, \frac{\partial V}{\partial y}=0$ and $\frac{\partial V}{\partial z}=0$
Hence $\vec{E}=-8 x \hat{i}$, so at point $(1 \mathrm{~m}, 0,2 \mathrm{~m})$
$\vec{E}=-8 \hat{i}$ volt/metre or 8 along negative $X$-axis.

Topic: Physics, Electrostatics, JEE Main
The electric potential $V$ at any point $(x, y, z)$, all in metres in space is given by $V=4 x^2$ volt. The electric field at the point $(1,0,2)$ in volt/metre is

Options:
A 8 along positive $X$-axis
B 16 along negative $X$-axis
C 16 along positive $X$-axis
D 8 along negative $X$-axis

The correct answer is:
8 along negative $X$-axis

Solution:
We know that
$\mathrm{E}=-\left[\hat{i} \frac{\partial V}{\partial x}+\hat{j} \frac{\partial V}{\partial y}+\mathrm{k} \frac{\partial V}{\partial z}\right]$
So
$\begin{aligned}
& \mathrm{E}=-\hat{i} \frac{\partial V}{\partial x}=-\hat{i} \frac{\partial}{\partial x}\left(4 x^2\right) \\
&=-8 x \hat{i} \mathrm{Vm}^{-1} \\
& \mathrm{E}_{(1,0,2)}=-8 \hat{i} \mathrm{Vm}^{-1}
\end{aligned}$

Topic: Physics, Electrostatics, NEET, NEET 2011 (Mains)
The electric potential $V$ is given as a function of distance $\times($ metre $)$ by $V=\left(5 x^{2}+10 x-4\right)$ volt.

Question: Value of electric field at $x=1 \mathrm{~m}$ is

Options:
A $-23 \mathrm{~V} / \mathrm{m}$
B $11 \mathrm{~V} / \mathrm{m}$
C $6 \mathrm{~V} / \mathrm{m}$
D $-20 \mathrm{~V} / \mathrm{m}$

The correct answer is:
$-20 \mathrm{~V} / \mathrm{m}$

Solution:
$\mathrm{V}=5 x^{2}+10 x-4$

$\mathrm{E}=\frac{-\mathrm{dV}}{\mathrm{dx}}=-(10 x+10)$

At $x=1 m, \mathrm{E}=-20 \mathrm{~V} / m$

Topic: Physics, Electrostatics, JEE Main
The electric potential $V$ is given as a function of distance $x$ (metre) by $V=\left(5 x^2+10 x-9\right)$ volt . Value of electric field at $x=1$ is

Options:
A $20 \mathrm{~V} / \mathrm{m}$
B $6 \mathrm{~V} / \mathrm{m}$
C $11 \mathrm{~V} / \mathrm{m}$
D $-23 V / m$

The correct answer is:
$20 \mathrm{~V} / \mathrm{m}$

Solution:
$E=-\frac{d V}{d x}=-\frac{d}{d x}\left(5 x^2+10 x-9\right)=-10 x-10$
$\therefore(E)_{x-1}=-10 \times 1-10=-20 \mathrm{~V} / \mathrm{m}$

Topic: Physics, Electrostatics, JEE Main
The electric potential $V(x)$ in a region around the origin is given by $V(x)=4 x^2$ volts. The electric charge enclosed in a cube of $1 \mathrm{~m}$ side with its centre at the origin is (in coulomb)

Options:
A
$8 \varepsilon_0$

B
$-4 \varepsilon_0$

C
0

D
$-8 \varepsilon_0$

The correct answer is:
0

Solution:
Charges reside only on the outer surface of a conductor with cavity.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2012 (07 May Online)
The electric resistance of a certain wire of iron is $R$. If its length and radius are both doubled, then

Options:
A The resistance will be doubled and the specific resistance will be halved.
B The resistance will be halved and the specific resistance will remain unchanged.
C The resistance will be halved and the specific resistance will be doubled.
D The resistance and the specific resistance, both will remain unchanged.

The correct answer is:
The resistance will be halved and the specific resistance will remain unchanged.

Solution:
$$
\begin{aligned}
& \text { (b) } \mathrm{R}^{\prime}=\frac{\rho \mathrm{L}}{\mathrm{A}} \\
& =\frac{\rho \times 2 \mathrm{~L}}{\pi(2 \mathrm{r})^2}=\frac{2 \rho \mathrm{L}}{\pi \times 4 \mathrm{r}^2}=\frac{\mathrm{R}}{2}
\end{aligned}
$$
Hence, the resistance will be halved and the specific resistance will remain unchanged.

Topic: Physics, Current Electricity, TS EAMCET, TS EAMCET 2023 (13 May Shift 1)
The electric resistance of a certain wire of iron is $R$. If its length and radius are both doubled, then:

Options:
A the resistance will be doubled and the specific resistance will be halved
B the resistance will be halved and the specific resistance will remain unchanged
C the resistance will be halved and the specific resistance will be doubled
D the resistance and the specific resistance, both will remain unchanged

The correct answer is:
the resistance will be halved and the specific resistance will remain unchanged

Solution:
Resistance of wire $R=\rho \frac{l}{A}$ Means $R \propto \frac{l}{A}=\frac{l}{\pi r^2}$
According to the question when $l$ and $R$ are doubled
$$
\begin{aligned}
& \therefore R_1 \propto \frac{2 l}{\pi(2 r)^2} \\
& \Rightarrow R_1 \propto \frac{1}{2} R
\end{aligned}
$$
Here specific resistance of wire is independent of geometry wire so it depends on the material of wire.

Topic: Physics, Current Electricity, NEET, NEET 2004
The electric resistance of a certain wire of iron is $R$. If its length and radius are both doubled, then

Options:
A the resistance and the specific resistance, will both remain unchanged
B the resistance will be doubled and the specific resistance will be halved
C the resistance will be halved and the specific resistance will remain unchanged
D the resistance will be halved and the specific resistance will be doubled

The correct answer is:
the resistance will be halved and the specific resistance will remain unchanged

Solution:
$\mathrm{R}_{1}=\frac{\rho \ell_{1}}{\mathrm{~A}_{1}}$, now $\ell_{2}=2 \ell_{1}$ and $\mathrm{r}_{2}=2 \mathrm{r}_{1}$
$$
\begin{array}{l}
\mathrm{A}_{1}=\pi \mathrm{r}_{1}^{2} \\
\mathrm{~A}_{2}=\pi\left(\mathrm{r}_{2}\right)^{2}=\pi\left(2 \mathrm{r}_{1}\right)^{2}=4 \pi \mathrm{r}_{1}^{2}=4 \mathrm{~A}_{1} \\
\mathrm{R}_{2}=\frac{\rho \ell_{2}}{\mathrm{~A}_{2}} \\
\therefore \quad R_{2}=\frac{\rho\left(2 \ell_{1}\right)}{4 A_{1}}=\frac{\rho \ell_{1}}{2 A_{1}}=\frac{R_{1}}{2}
\end{array}
$$
$\therefore$ Resistance is halved, but specific resistance remains the same.

Topic: Physics, Current Electricity, VITEEE, VITEEE 2019
The electrical circuit used to get smooth dc output from a rectifier circuit is called

Options:
A Oscillator
B Filter
C Amplifier
D Logic gates

The correct answer is:
Filter

Solution:
Filter circuits are used to get smooth dc $\pi$ filter is the best filter.

Topic: Physics, Semiconductors, JEE Main
The electrical conductivity of a semiconductor increases when a radiation of wavelength shorter than $1000 \mathrm{~nm}$ is incident on it. The band gap of the semiconductor is:

Options:
A $2.4 \mathrm{eV}$
B $1.2 \mathrm{eV}$
C $3.4 \mathrm{eV}$
D $4.0 \mathrm{eV}$

The correct answer is:
$1.2 \mathrm{eV}$

Solution:
Concept:
The freq. and wavelength are related by the relation:
$f=\frac{C}{2}$
Also, the energy of a photon with wavelength $(\lambda)$ is given by:
$E=\frac{h c}{\lambda} J$
Where,
h: Planck's constant $=6.626 \times 10-34 \mathrm{Js}$
c: $3 \times 108 \mathrm{~m} / \mathrm{sec}$
$\begin{aligned}
& E=\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{\lambda} \text { joule } \\
& \because 1 \mathrm{eV}=1.602 \times 10^{-19} \text { Joule } \\
& E=\frac{19.878 \times 10^{-26}}{1.002 \times 10^{-15} \lambda} \mathrm{ev} \\
& E \approx \frac{12400}{\lambda(\lambda)} \mathrm{eV}\ldots(1)
\end{aligned}$
Calculation:
Given:
$\lambda=1000 \mathrm{~nm}=1 \mu \mathrm{m}$
From equation (1)
$\begin{aligned}
& E=\frac{12400}{1 \mu m} \\
& E=1.24 \mathrm{eV}
\end{aligned}$

Topic: Physics, Semiconductors, JEE Main
The electrical conductivity of a semiconductor increases when electromagnetic radiation of a wavelength shorter than $2480 \mathrm{~nm}$ is incident on it The bandgap in $(\mathrm{eV})$ for the semiconductor is

Options:
A $0.5 \mathrm{eV}$
B $0.7 \mathrm{eV}$
C $1.1 \mathrm{eV}$
D $2.5 \mathrm{eV}$

The correct answer is:
$0.5 \mathrm{eV}$

Solution:
Band gap $=$ Energy of photon of $=2480 \mathrm{~nm}$
Energy $=\left(h_c / \lambda\right) J=\left(h_c / \lambda_e\right) e V$
Band gap $=\left(\left[\left(6.63 \times 10^{-34}\right) \times\left(3 \times 10^8\right)\right] /\left[\left(2480 \times 10^{-9}\right) \times\left(1.6 \times 10^{-19}\right)\right]\right)=0.5 \mathrm{eV}$

Topic: Physics, Semiconductors, JEE Main
The electrical conductivity of a semiconductor increases when electromagnetic radiation of a wavelength shorter than 2480 nm is incident on it. The bandgap in (eV) for the semiconductor is [take hc=12400 eV A∘]

Options:
A 0.5 eV
B 0.7 eV
C 1.1 eV
D 2.5 eV

The correct answer is:
0.5 eV

Solution:
λ max = 2 4 8 0 nm = 2 4 8 0 0 Å
Energy (in eV) =12400 eV Åλ
E=1240024800 eV
E=0.5 eV

Topic: Physics, Semiconductors, JEE Main
The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than $2480 \mathrm{~nm}$ is incident on it. The band gap in (eV) for the semiconductor is

Options:
A
$1.1 \mathrm{eV}$

B
$2.5 \mathrm{eV}$

C
$0.5 \mathrm{eV}$

D
$0.7 \mathrm{eV}$

The correct answer is:
$0.5 \mathrm{eV}$

Solution:
$\mathrm{E}_{\mathrm{g}}=\frac{\mathrm{hc}}{\lambda}=0.5 \mathrm{eV}$

Topic: Physics, Semiconductors, JEE Main, JEE Main 2005
The electrical resistance $R$ of a conductor of length $l$ and area of cross section $a$ is given by $R=\frac{\rho l}{a}$ where ' $\rho$ ' is the electrical resistivity. What is the dimensional formula for electrical conductivity ' $\sigma$ ' which is reciprocal of resistivity?

Options:
A
$\left[M^{-1} L^{-3} T^3 A^2\right]$

B
$\left[M L^{-3} T^{-3} A^2\right]$

C
$\left[M L^3 T^{-3} A^{-2}\right]$

D
$\left[M^{-2} L^3 T^2 A^{-1}\right]$

The correct answer is:
$\left[M^{-1} L^{-3} T^3 A^2\right]$

Solution:
Dimensional formula for electrical resistivity, $\rho=\left[\mathrm{ML}^3 \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]$ Dimensional formula for electrical conductivity, $\sigma=\frac{1}{\rho}$ is $\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^3 \mathrm{~A}^2\right]$

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2012 (26 May Online)
The electrochemical equivalent of a metal is $3.3 \times 10^{-7} \mathrm{~kg}$ per coulomb. The mass of the metal liberated at the cathode when a 3 A current is passed for 2 seconds will be

Options:
A
$19.8 \times 10^{-7} \mathrm{~kg}$

B
$9.9 \times 10^{-7} \mathrm{~kg}$

C
$6.6 \times 10^{-7} \mathrm{~kg}$

D
$1.1 \times 10^{-7} \mathrm{~kg}$

The correct answer is:
$19.8 \times 10^{-7} \mathrm{~kg}$

Solution:
$\mathrm{m}=$ Zit,
$\mathrm{m}=3.3 \times 10^{-7} \times 3 \times 2=19.8 \times 10^{-7} \mathrm{~kg}$.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2004
The electrode potential, E° , for the reduction of MnO4- to Mn2+ in acidic medium is +1.51 V. Which of the following metal(s) will be oxidised? The reduction reactions and standard electrode potentials for Zn2+,Ag+ and Au+ are given as

Znaq2++2e-→Zns,E°=-0.762V

Agaq++e-⇌Agx,E°=+0.80V

Auaq++e-⇌Aus,E°=+1.69V

Options:
A Zn and Au
B Ag and Au
C Au
D Zn and Ag

The correct answer is:
Zn and Ag

Solution:
More +ve the value of reduction potential more will be the tendency to undergo reduction.
The E red o MnO 4 − | Mn 2+ =+1.51 V value is greater than E red o Zn 2+ |Zn =−0.762 V and E red o Ag + |Ag =+0.80 V. Thus, only Zn and Ag will be oxidised.

Topic: Chemistry, Redox Reactions, JEE Main
The electrode potential of M2+/M of 3 d-series elements shows positive value for?

Options:
A Co
B Fe
C Zn
D Cu

The correct answer is:
Cu

Solution:
Only copper shows positive value for electrode potential of M2+/M of 3 d-series elements.

EΘCu2+/Cu=+0.34

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2021 (24 Feb Shift 1)
The electrode potential of the following half cell at 298 K

XX2+0.001M‖Y2+0.01MY is _____ ×10-2 V (Nearest integer)

Given : EX2+∣X0=-2.36 V

EY2+Y0=+0.36 V

2.303RTF=0.06 V

Options:

The correct answer is:
275

Solution:
The net cell reaction can be written as follows,

X+Y2+→Y+X2+

Ecell0=Ecathode0-Eanode0(Both are SRP values)

ECell0=0.36--2.36=2.72 V

Now, the Nernst equation is,

Ecell=Ecell0-0.06nlogX2+Y2+

n=The number of electrons involved in the net reaction.

ECell=2.72-0.062log0.0010.01

=2.72+0.03=2.75 V

=275×10-2 V

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2023 (30 Jan Shift 2)
The electrolytes usually used in the electroplating of gold and silver, respectively, are:

Options:
A AuNH32+ and AgCN2-
B AuCN2- and AgCl2-
C AuOH4- and AgOH2-
D AuCN2- and AgCN2-

The correct answer is:
AuCN2- and AgCN2-

Solution:
The solution of dicyano complexes of gold (Au) and silver (Ag) are used as electrolytes while performing electroplating with them.

The amine complexes of gold are not stable, so, they cannot be used.

Silver cyanide is employed as cyanide reacts with silver to form stable complexes.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2019 (10 Jan Shift 2)
The electrolytic method of reduction is employed for the preparation of metals that

Options:
A are weakly electropositive
B are moderately electropositive
C are strongly electropositive
D form acidic oxides

The correct answer is:
are strongly electropositive

Solution:
Pure metal is obtained at cathode and impurities are obtained at anode.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
The electromagnetic waves travel in free space with the velocity of

Options:
A sound
B light
C greater than that of light
D greater than that of sound

The correct answer is:
light

Solution:
Electromagnetic waves travel in free space or vacuum with the velocity of light $\left(3 \times 10^{8} \mathrm{~ms}^{-1}\right)$.

Topic: Physics, Electromagnetic Waves, JEE Main
The electron affinity of chlorine is $3.7 \mathrm{eV} .1$ gram of chlorine is completely converted to $\mathrm{Cl}^{-}$ion in a gaseous state. $\left(1 \mathrm{eV}=23.06 \mathrm{kcal} \mathrm{mol}^{-1}\right)$. Energy released in the process is

Options:
A
$4.8 \mathrm{kcal}$

B
$7.2 \mathrm{kcal}$

C
$8.2 \mathrm{kcal}$

D
$2.4 \mathrm{kcal}$

The correct answer is:
$2.4 \mathrm{kcal}$

Solution:
Number of moles $=\frac{1}{35.5}$
Given, $1 \mathrm{eV}=23.06 \mathrm{kcal} \mathrm{mol}^{-1}$
$3.7 \mathrm{eV}=3.7 \times 23.06 \mathrm{kcal} \mathrm{mol}^{-1}$
i.e. 1 mole realease energy $=3.7 \times 23.06 \mathrm{kcal}$
$\therefore$ Energy released
$$
=\frac{1}{35.5} \times 3.7 \times 23.06 \mathrm{kcal}=2.4 \mathrm{kcal}
$$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2012 (07 May Online)
The electron affinity value are negative for
A. \(\mathrm{Be} \rightarrow \mathrm{Be}^{-}\)
B. \(\mathrm{N} \rightarrow \mathrm{N}^{-}\)
C. \(\mathrm{O} \rightarrow \mathrm{O}^{2-}\)
D. \(\mathrm{Na} \rightarrow \mathrm{Na}^{-}\)
E. \(\mathrm{Al} \rightarrow \mathrm{Al}^{-}\)
Choose the most appropriate answer from the options given below :

Options:
A D and E only
B A, B and C only
C A and D only
D A, B, D and E only

The correct answer is:
A, B and C only

Solution:
(A) $\mathrm{Be}+\mathrm{e}^{-} \rightarrow \mathrm{Be}^{-}, \quad$ E.A $=-$ ive
(B) $\mathrm{N}+\mathrm{e}^{-} \rightarrow \mathrm{N}^{-} \quad$ E.A $=$-ive
(C) $\mathrm{O}+\mathrm{e}^{-} \rightarrow \mathrm{O}^{-}$
$\mathrm{O}^{-}+\mathrm{e}^{-} \rightarrow \mathrm{O}^{-2} \quad$ E.A $=-$ ive
(D) $\mathrm{Na}+\mathrm{e}^{-} \rightarrow \mathrm{Na}^{-} \quad$ E.A $=+$ ive
(E) $\mathrm{A} \ell+\mathrm{e}^{-} \rightarrow \mathrm{A} \ell^{-} \quad$ E.A $=+$ ive

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2024 (06 Apr Shift 1)
The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths λ1λ2 of the photons emitted in this process is:

Options:
A 75
B 207
C 97
D 275

The correct answer is:
207

Solution:
1λ1=R132-142= R×7144
1λ2=R122-132= R×536
⇒λ1λ2=R×536×144R×7=207

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2019 (12 Apr Shift 2)
The electron in a hydrogen atom makes a transition from a state n1 to n2. If the time period of the electron in the initial state is eight times that in the final state, then the possible values of n1 and n2 are

Options:
A n1=8, n2=2
B n1=4, n2=2
C n1=3, n2=1
D n1=4, n2=1

The correct answer is:
n1=4, n2=2

Solution:
In a hydrogen atom, the time period obeys the following relation

T∝n3
⇒T1T2=n1n23

⇒81=n1n23

⇒n1n2=21
Thus, n1=4 and n2=2

Topic: Physics, Atomic Physics, JEE Main
The electron in a hydrogen atom makes a transition from $n=n_{1}$ to $n=n_{2}$ state. The time period of the electron in the initial state $n_{1}$ is eight times that in the final state $n_{2}$. The possible values of $n_{1}$ and $n_{2}$ are

Options:
A $n_{1}=8, n_{2}=1$
B $n_{1}=4, n_{2}=2$
C $n_{1}=2, n_{2}=4$
D $n_{1}=1, n_{2}=8$

The correct answer is:
$n_{1}=4, n_{2}=2$

Solution:
Here, for hydrogen atom, $Z=1$
We know that, time period of electron in $n$th orbit is given as
$\begin{aligned} T_{n} & \propto n^{3} \\ \therefore \quad & \frac{T_{n_{1}}}{T_{n_{2}}}=\frac{n_{1}^{3}}{n_{2}^{3}} \end{aligned}$
Since, $\quad T_{n_{1}}=8 T_{n_{2}}$
$\therefore \quad \frac{8 T_{n_{2}}}{T_{n_{2}}}=\frac{n_{1}^{3}}{n_{2}^{3}} \Rightarrow 2^{3}=\left(\frac{n_{1}}{n_{2}}\right)^{3}$
$\Rightarrow \quad n_{1}=2 n_{2}$
Hence, this condition satisfies only in option (b).
i.e., $\quad n_{1}=4$ and $n_{2}=2$

Topic: Physics, Atomic Physics, JEE Main
The electron in the hydrogen atom jumps from excited state $(n=3)$ to its ground state $(n=1)$ and the photons thus emitted irradiate a photosensitive material. If the work function of the material is $5.1 \mathrm{eV}$, the stopping potential is estimated to be (the energy of the electron in $n$th state $\mathrm{E}_{\mathrm{n}}=-\frac{13.6}{\mathrm{n}^2} \mathrm{eV}$ )

Options:
A $5.1 \mathrm{~V}$
B $12.1 \mathrm{~V}$
C $17.2 \mathrm{~V}$
D $7 \mathrm{~V}$

The correct answer is:
$7 \mathrm{~V}$

Solution:
For $n=1, \mathrm{E}_1=-\frac{13.6}{(1)^2}=-13.6 \mathrm{eV}$ and for $n=3, E_3=-\frac{13.6}{(3)^2}=-1.51 \mathrm{eV}$
So, required energy
$\begin{array}{l}
\mathrm{E}=\mathrm{E}_3-\mathrm{E}_1 =-(1.51)-(-13.6) =12.09 \mathrm{eV} \\
\because \mathrm{E} =\mathrm{W}+\mathrm{eV} \\
\therefore \mathrm{eV} =\mathrm{E}-\mathrm{W} \\
\mathrm{eV} =(12.09-51) \mathrm{e} =7 \text { volt }
\end{array}$

Topic: Physics, Atomic Physics, NEET, NEET 2010 (Mains)
The electron in the hydrogen atom undergoes transition from higher orbitals to orbital of radius 211.6 pm. This transition is associated with

Options:
A Lyman series
B Balmer series
C Brackett series
D Paschen series

The correct answer is:
Balmer series

Solution:
R=211.6 pm=2.11Å.
R=0.529×n2Z=2.11Å.
n2=4⇒n=2.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2017 (09 Apr Online)
The electron in the nth orbit of Li2+ is excited to (n+1) orbit using the radiation of energy 1.47×10-17 J (as shown in the diagram). The value of n is ___________

Given : RH=2.18×10-18 J

Options:

The correct answer is:
1

Solution:
The difference energy between two energy levels is

∆E=2.18×10-18 1n12-1n22 J

1.47×10-17=2.18×10-18×(9)1(n)2-1(n+1)2

0.749=1n2-1(n+1)2

n = 1

Hence, the transition is taking place between 1→2, and it comes under Lymann series.

If we take 0.749≃0.75=34

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2023 (10 Apr Shift 2)
The electron of a hydrogen atom makes a transition from the $(n+1)^{\text {th }}$ orbit to the $n^{\text {th }}$ orbit. For large $\mathrm{n}$ the wavelength of the emitted radiation is proportional to

Options:
A
$n$

B
$n^3$

C
$n^4$

D
$n^2$

The correct answer is:
$n^3$

Solution:
If $n_1=n$ and $n_2=n+1$ Maximum wavelength $\lambda_{\max }=\frac{n^2\left(n+1^2\right)}{(2 n+1 R)}$
Therefore, for large $\mathrm{n}, \lambda_{\max } \propto n^3$

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2012 (07 May Online)
The electronegativity of aluminum is similar to:

Options:
A Beryllium
B Boron
C Carbon
D Lithium

The correct answer is:
Beryllium

Solution:
EN of Al=1.61
EN of Be=1.57
EN of B=2.04
EN of C=2.55
Due to diagonal relationship between Be and Al, they exhibit similar properties and thus, have similar electronegativity values.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2019 (10 Jan Shift 1)
The electronic configuration for Neodymium is:

[Atomic Number for Neodymium 60]

Options:
A [Xe] 4f4 6s2
B [Xe] 5f47s2
C [Xe] 4f6 6s2
D [Xe] 4f15d16s2

The correct answer is:
[Xe] 4f4 6s2

Solution:
The atomic number of neodymium is 60.

Therefore, the electronic configuration of neodymium (Nd) can be determined by distributing its electrons according to the Aufbau principle, Pauli exclusion principle, and Hund's rule.

Electronic configuration of Nd(Z = 60) is;

[Xe] 4f46s2

as the last electron of Neodymium enters the f subshell. so it is called a f-block element.

Hence option A is the answer.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2024 (27 Jan Shift 1)
The electronic configuration of atomic number 20 of an atom is which of the following?

Options:
A $2,6,6,2$
B $2,8,8,2$
C $2,4,6,2$
D $2,4,6,2$

The correct answer is:
$2,8,8,2$

Solution:
Atomic number of calcium is 20 . Number of protons $=$ number of electrons $=$ 20
First shell can accommodate 2 electrons so $\mathrm{K}=2$.
Second shell can accommodate maximum of 8 electrons so $\mathrm{L}=8$.
Third shell can accommodate maximum of 18 electrons but if we put remaining 10 electrons it will violate the law that outermost electron can accommodate 8 electrons so it will accommodate only 8 electrons. So number of electrons in $\mathrm{M}$ is 8 . Now the remaining two will be in $\mathrm{N}$. So the configuration is $2,8,8,2$

Topic: Physics, Atomic Physics, JEE Main
The electronic configuration of chromium $(z=24)$ is:

Options:
A $[\mathrm{Ne}] 3 s^2 3 p^6 3 d^4 4 s^2$
B $[\mathrm{Ne}] 3 s^2 3 p^6 3 d^5 4 s^1$
C $[\mathrm{Ne}] 3 s^2 3 p^6 3 d^1 4 s^2$
D $[\mathrm{Ne}] 3 s^2 3 p^6 4 s^2 4 p^4$

The correct answer is:
$[\mathrm{Ne}] 3 s^2 3 p^6 3 d^5 4 s^1$

Solution:
The electronic configuration of chromium $(Z=24)$ is $[N e] 3 s^2 3 p^6 3 d^5 4 s^1$.

Topic: Physics, Atomic Physics, JEE Main
The electronic configuration of cobalt is

Options:
A $1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^1, 4 s^2$
B $1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^7, 4 s^2$
C $1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^3, 4 s^2$
D $1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^5, 4 s^2$

The correct answer is:
$1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^7, 4 s^2$

Solution:
The electron configuration of an atom specifies the arrangement of electrons in the various orbitals around the nucleus. For cobalt (Co), which has 27 electrons, the configuration is written as $1 s^2 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^7, 4 s^2$.
In this notation, the number before the orbital symbol specifies the principal energy level, and the superscript represents the number of electrons in that orbital. So, for example, the notation $1 s^2$ indicates that there are two electrons in the first principal energy level ( $n=$ 1) in an s orbital.
The Aufbau principle dictates how electrons fill orbitals in atoms. Electrons fill orbitals from lowest energy to highest energy. The order of filling is: $1 s, 2 s, 2 p, 3 s, 3 p, 4 s, 3 d, 4 p, 5 s, 4 d, 5 p$, $6 s, 4 f, 5 d, 6 p, 7 s, 5 f, 6 d$, and $7 p$. There are exceptions to this order for some elements due to the stability of half-filled and fully filled orbitals. Cobalt is an example of this.
According to the Aufbau principle, the $4 \mathrm{~s}$ orbital should fill before the $3 \mathrm{~d}$ orbitals. However, filling the 3d orbitals with seven electrons creates a more stable configuration for cobalt. This is because the $3 \mathrm{~d}$ and $4 \mathrm{~s}$ orbitals have very similar energies, and the extra electron in the $3 \mathrm{~d}$ orbital creates a halffilled subshell, which is energetically favorable. So the electron configuration for cobalt is $1 s^2 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^7, 4 s^2$.

Topic: Chemistry, d and f Block Elements, JEE Main