Radiations of two photons having energies twice and five times the work function of a metal are incident successively on a metal surface. The ratio of the maximum velocity of photoelectrons emitted in the two cases will be

  • A
    $1: 1$
  • B
    $1: 2$
  • C
    $1: 3$
  • D
    $1: 4$

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When a metal is irradiated by light having wavelength $\lambda$ $(\lambda < \lambda_0)$,all the photoelectrons emitted are bent in a circle of radius $r$ by a magnetic field of flux density $B_0$. Find $\frac{1}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength.

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The photoelectric threshold wavelength for potassium (work function being $2 \ eV$) is ........... $nm$.

$A$ photoemissive substance is illuminated with a radiation of wavelength $\lambda_i$ so that it releases electrons with de-Broglie wavelength $\lambda_e$. The longest wavelength of radiation that can emit photoelectron is $\lambda_0$. The expression for the de-Broglie wavelength $\lambda_e$ is ($m = \text{mass of electron}$, $h = \text{Planck's constant}$, $c = \text{speed of light}$):

Two photons of energies twice and thrice the work function of a metal are incident on the metal surface. Then,the ratio of maximum velocities of the photoelectrons emitted in the two cases respectively,is

$A$ beam of light falls on a metal surface such that photo-electrons are generated. If the power of the light source starts to decrease linearly with time $t$, then the variation of the photocurrent $I$ and the magnitude of the stopping potential $|V|$ with time is best represented by:

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