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. $7 \mathrm{V}$

In the phenomenon of electric discharge through gases at low pressure, the coloured glow in the tube appears as a result of:

1. excitation of electrons in the atoms 

2. the collision between the atoms of the gas 

3. the collisions between the charged particles emitted from the cathode and the atoms of the gas 

4. the collision between different electrons of the atoms of the gas 

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) $4\to 3$

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

Electrons of mass m with de- Broglie wavelength λ fall on the target in an X-ray tube. The cut-off wavelength (λ_o) of the emitted X-ray is:

1.  ${\mathrm{\lambda }}_0=\frac{2{\mathrm{mc\lambda }}^2}{\mathrm{h}}$

2.  ${\mathrm{\lambda }}_0=\frac{2\mathrm{h}}{\mathrm{mc}}$

3.  ${\mathrm{\lambda }}_0=\frac{2{\mathrm{m}}^2{\mathrm{c}}^2{\mathrm{\lambda }}^2}{{\mathrm{h}}^2}$

4.  ${\mathrm{\lambda }}_0=\mathrm{\lambda }$

The ratio of wavelengths of the last line of the Balmer series and the last line of the Lyman series is:
1. \(1\)
2. \(4\)
3. \(0.5\)
4. \(2\)