According to Einstein's photoelectric equation, the graph of KE of the photoelectron emitted from the metal versus the frequency of the incident radiation gives a straight line graph, whose slope

• depends on the intensity of the incident radiation

• depends on the nature of the metal and also on the intensity of incident radiation

• is same for all metals and independent of the intensity of the incident radiation

• depends on the nature of the metal

The photoelectric threshold wavelength for silver is M. The energy of the electron ejected from the surface of silver by an incident wavelength λ (λ < λ₀) will be

• hc (λ₀ − λ)

• $\frac{\mathrm{hc}}{{\mathrm{\lambda }}_0 - \mathrm{\lambda }}$

• $\frac{\mathrm{h}}{\mathrm{c}} \left(\frac{{\mathrm{\lambda }}_0 - \mathrm{\lambda }}{{\mathrm{\lambda \lambda }}_0}\right)$

• $\mathrm{hc} \left(\frac{{\mathrm{\lambda }}_0 - \mathrm{\lambda }}{{\mathrm{\lambda \lambda }}_0}\right)$

When an electron jumps from the orbit n = 2 to n = 4, then wavelength of the radiations absorbed will be (R is Rydberg's constant)

• $\frac{16}{3\mathrm{R}}$

• $\frac{16}{5\mathrm{R}}$

• $\frac{5\mathrm{R}}{16}$

• $\frac{3\mathrm{R}}{16}$

n photons of wavelength λ are absorbed by a black body of mass m. The momentum gained by the body is

• $\frac{\mathrm{nh}}{\mathrm{\lambda }}$

• $\frac{\mathrm{h}}{\mathrm{m\lambda }}$

• $\frac{\mathrm{mnh}}{\mathrm{\lambda }}$

• $\frac{\mathrm{nh}}{\mathrm{m\lambda }}$

55.

Light emitted during the de excitation of electron from n = 3 to n = 2, when incident on a metal, photoelectrons are just emitted from that metal. In which of the following de excitations photoelectric effect is not possible ?

• From n = 2 to n = 1

• From n = 3 to n = 1

• From n = 5 to n = 2

• From n = 4 to n = 3

Maximum velocity of the photoelectron emitted by a metal is 1.8 x 10⁶ ms^-1. Take the value of specific charge of the electron is 1.8 x 10¹¹ C kg^-1. Then the stopping potential in volt is

• 1

• 3

• 9

• 6

The maximum kinetic energy of the photoelectrons depends only on

• potential 

• frequency

• incident angle

• pressure

Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively, successively illuminate a metallic surface whose work function is 0.5 eV. Ratio of maximum speeds of emitted electrons will be

• 1 : 2

• 1 : 5

• 1 : 1

• 1 : 4

The variation of photocurrent with collector potential for different frequencies of incident radiation v₁, v₂ and v₃ is as shown in the graph, then

• v₁ = v₂ = v₃

• v₁ > v₂ > v₃

• v₁ < v₂ < v₃

• ${\mathrm{v}}_3 = \frac{{\mathrm{v}}_1 + {\mathrm{v}}_2}{2}$

From the following graph of photo current against collector plate potential, for two different intensities of light I₁ and I₂, one can conclude

• I₁ = I₂

• I₁ > I₂

• I₁ < I₂

• Comparision is not possible