Balmer series contains wavelengths ranging from 364 nm (for n2​=3 ) to 655 nm(n2​=∞). So, the given range of wavelength (380−780 nm) lies in the Balmer series. The wavelength in the Balmer series can be found by​λ1​=R(221​−n21​) Here, R= Rydberg’s constant =1.097×107 m−1​ The wavelength for the transition from n=3 to n=2 is given by​λ1​1​=R(221​−321​)λ1​=656.3 nm​ The wavelength for the transition from n=4 to n=2 is given by​λ2​1​=R(221​−421​)λ2​=486.1 nm​ The wavelength for the transition from n=5 to n=2 is given by​λ3​1​=R(221​−521​)λ3​=434.0 nm​ The wavelength for the transition from n=6 to n=2 is given byλ4​1​λ4​​=R(221​−621​)=410.2 nm​ The wavelength for the transition from n=7 to n=2 is given by​λ5​1​=R(221​−721​)λ5​=397.0 nm​ Thus, the wavelengths emitted by the atomic hydrogen in visible range (380-780 nm ) are 5 .Lyman series contains wavelengths ranging from 91 nm (for n2​=2) to 121  nm(n2​=∞). So, the wavelengths in the given range (50-100 nm) must lie in the Lyman series. The wavelength in the Lyman series can be found byλ1​=R(121​−n21​) The wavelength for the transition from n=2 to n=1 is given by​λ1​1​=R(121​−221​)λ1​=122 nm​ The wavelength for the transition from n=3 to n=1 is given by​λ2​1​=R(121​−221​)λ2​=103 nm​ The wavelength for the transition from n=4 to n=1 is given by​λ3​1​=R(121​−421​)λ3​=97.3 nm​ The wavelength for the transition from n=5 to n=1 is given by​λ4​1​=R(121​−521​)λ4​=95.0 nm​ The wavelength for the transition from n=6 to n=1 is given by​λ5​1​=R(121​−621​)λ5​=93.8 nm​ So, it can be noted that the number of wavelengths lying between 50  nm to 100  nm are 3 .