MHT CET 2019 3rd May Morning Shift | Dual Nature of Radiation Question 85 | Physics

When certain metal surface is illuminated with a light of wavelength $\lambda$, the stopping potential is $V$, When the same surface is illuminated by light of wavelength $2 \lambda$, the stopping potential is $\left(\frac{V}{3}\right)$. The threshold wavelength for the surface is

$\frac{8 \lambda}{3}$

$\frac{4 \lambda}{3}$

$4 \lambda$

$6 \lambda$

MHT CET 2019 2nd May Evening Shift

MCQ (Single Correct Answer)

-0

A metal surface is illuminated by light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one fourth of its original value then the maximum KE of the emitted photoelectrons would be

twice the original value

four times the original value

one fourth of the original value

unchanged

MHT CET 2019 2nd May Morning Shift

MCQ (Single Correct Answer)

-0

When photons of energy $h v$ fall on a metal plate of work function ' $W_0$ ', photoelectrons of maximum kinetic energy ' $K$ ' are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be

$K+W_0$

$K+h v$

$\mathrm{K}$

$\mathrm{2 K}$

MHT CET 2019 2nd May Morning Shift

MCQ (Single Correct Answer)

-0

The maximum velocity of the photoelectron emitted by the metal surface is ' $v$ '. Charge and mass of the photoelectron is denoted by ' $e$ ' and ' $m$ ' respectively. The stopping potential in volt is

$\frac{v^2}{2\left(\frac{m}{e}\right)}$

$\frac{v^2}{2\left(\frac{e}{m}\right)}$

$\frac{v^2}{\left(\frac{e}{m}\right)}$

$\frac{v^2}{\left(\frac{m}{e}\right)}$

MHT CET Subjects