Given the value of Rydberg constant is ${10}^7 m^{-1}$, the wave number of the last line of the Balmer series in hydrogen spectrum will be:

(a) $0.5\times {10}^7{m}^{-1}$        (b) $0.25\times {10}^7{m}^{-1}$

(c) $2.5\times {10}^7{m}^{-1}$        (d) $0.025\times {10}^4{m}^{-1}$

Electron in hydrogen atom first jumps from third exicted state to second exicted state and then from second exicted to the first excited state. The ratio of the wavelengths ${\lambda }_1:{\lambda }_2$ emitted in the two cases is

(1) 7/5

(2) 27/20

(3) 27/5

(4) 20/7

An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquires as a result of photon emission will be:
(m is the mass of the hydrogen atom, \(R\) Rydberg constant and \(h\) Planck's constant)

The transition from the state n=3 to n=1

a hydrogen like atom results in ultraviolet

radiation. Infrared radiation will be obtained

in the transition from

(a) $2\to 1$

(b) $3\to 2$

(c) $4\to 2$

(d) $4\to 3$

The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen-like ion. The atomic number Z of hydrogen-like ion is: 

(1) 4                                                 

(2) 1

(3) 2                                                 

(4) 3

An electron in the hydrogen atom jumps from nth excited state to the ground state. The wavelength so emitted illuminates a photosensitive material having work function 2.75 eV. If the stopping potential of the photo-electron is 10V, the value of n is 

(1) 3

(2) 4

(3) 5

(4) 2

An alpha nucleus of energy $\frac{1}{2}m{v}^2$ bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to

(a) $\frac{1}{Ze}$                              (b) $v^2$

(c) $\frac{1}{m}$                               (d) $\frac{1}{v^4}$