English: So, from the first condition ∫p_(ϕ)dϕ = n_(ϕ)h, I get L=  n_(ϕ)ℏ and from the second condition that ∫p_(r)dr= n_(r)h, I get $- 2\pi\left( L - \frac{mk}{\ } \right)\ = n_{r}h$ and therefore, $L - \frac{mk}{\ } = - n_{r}\hslash$ with the minus sign or L which is m_(ϕ)ℏ, if I substitute that I get $\left( n_{\phi} + n_{r} \right)\hslash = \frac{mk}{\ }\ $or E equals n phi plus n r equals m k², there will be an amount of n²ℏ² with the minus sign which is same as $\frac{mZ^{2}e^{4}}{32\pi^{2}\epsilon_{0}^{2}\hslash^{2}}\ $to the minus $\left( n_{\phi} + n_{r} \right)\hslash = \frac{mk}{\ }$ ಆಗಿರುತ್ತದೆ E So, we getting an energy E which is equal to $\left( - \frac{mZ^{2}e^{4}}{32\pi^{2}\epsilon_{0}^{2}\hslash^{2}} \right)1/n^{2}\ $, let me call this n where n is n_(ϕ) + n_(r).
Kannada: ಆದ್ದರಿಂದ, ಮೊದಲ ಷರತ್ತಿನಿಂದ ∫p_(φ)dφ = n_(φ)h, ನನಗೆ L= ∫p_ n_(φ)(r)dr= n_(r)h, ನಾನು $- 2\pi\left(L - \frac{mk}{\ } \right)\ = n_{r}m_ h$ ಮತ್ತು ಆದ್ದರಿಂದ, $L φ - \frac{mk}{\ = n_n_  ನಾನು $\left(n_{\phi} + n_{r} \right)\hslash = \frac{mk}{\ }\ $or E ಅನ್ನು n phi plus n r m k² ಗೆ ಸಮನಾಗಿಸಿದರೆ, $to n² ನ ಪ್ರಮಾಣವು $\frac{mz^{2}e^{4}}{32\pi^n_ {2}\epsilon_{0} {2}\$to n_\hslash=hslash}\ $to=frac{mk}{\ {32\p^$to\hslash\{2}}\மருரரு்குவகருரருனனுபுருரருஷ்னுகுிᅗಉுிᅗരതಈுிᅗுᅗപ국궗പ궗೫국പᅗ국국പ국어어  ನಾವು $\left(- \frac{mZ^{2}e^{4}}{32\pi^{2}\epsilon_{0}^{2}\hslash^{2}} \right)1/n^{2}\ $ಗೆ ಸಮನಾದ E ಶಕ್ತಿಯನ್ನು ಪಡೆಯುತ್ತೇವೆ, ಇದನ್ನು n ಅನ್ನು n_ (φ) + n_(r) ಎಂದು ಕರೆಯುತ್ತೇನೆ.