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

52.

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

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

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

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

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

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