Considering the \(3^{rd}\) orbit of \(\mathrm{He}^{+}\) (Helium ion), using the non-relativistic approach, the speed of the electron in this orbit will be: (Given: \(Z=2, K = 9\times 10^{9}\), and Planck's constant, \(h= 6.6\times10^{-34}\) J-s)
1. \(2.92\times 10^{8}\) m/s
2. \(1.46\times 10^{6}\) m/s
3. \(0.73\times 10^{8}\) m/s
4. \(3.0\times 10^{8}\) m/s

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) $\frac{1}{v^4}$

The electron in the hydrogen atom jumps from excited state $\left(n=3\right)$ to its ground state $\left(n=1\right)$ 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 the nth state ${\mathrm{E}}_{\mathrm{n}}=-\frac{13.6}{{\mathrm{n}}^2}\mathrm{eV}$)

1. $5.1 \mathrm{V}$                                               

2. $12.1 \mathrm{V}$

3. $17.2 \mathrm{V}$                                             

4. $7 \mathrm{V}$