According to Einstein's photoelectric equatio | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Einstein's photoelectric equation is given by $K_{max} = h
u - \phi_0$,where $K_{max}$ is the maximum kinetic energy,$h$ is Planck's constant,$

Vedclass · Accepted Answer

is the same for all metals and is independent of the intensity of radiation.

Vedclass · Answer

(D) Einstein's photoelectric equation is given by $K_{max} = h
u - \phi_0$,where $K_{max}$ is the maximum kinetic energy,$h$ is Planck's constant,$
u$ is the frequency of incident radiation,and $\phi_0$ is the work function of the metal.Comparing this with the equation of a straight line $y = mx + c$,we get $K_{max} = h
u - \phi_0$.Here,the slope $m = h$ (Planck's constant).Since Planck's constant $h$ is a universal constant,the slope of the graph is the same for all metals.Furthermore,the slope is independent of the intensity of the incident radiation.

According to Einstein's photoelectric equation,the graph of the maximum kinetic energy of emitted photoelectrons versus the frequency of incident radiation is a straight line. Its slope . . . . . .

Similar Questions

Energy of the incident photons on the photosensitive metal surface is $3W$ and then $5W$ where '$W$' is the work function of that metal. The ratio of velocities of the emitted photoelectrons is

$V$ (stopping potential) is plotted against $\frac{1}{\lambda}$,where $\lambda$ is the wavelength of incident radiation,for two metals.

The work function of a metal is $3 \ eV$. The color of the visible light that is required to cause emission of photoelectrons is

When the energy of the incident radiation is increased by $20\%$,the kinetic energy of the photoelectrons emitted from a metal surface increases from $0.5\, eV$ to $0.8\, eV$. The work function of the metal is ............. $eV$.

The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface,having work function $5.01 \, eV$,when ultraviolet light of $200 \, nm$ falls on it,must be ............... $V$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module