In an experiment on photoelectric emission fr | vedclass

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

Question

(C) According to Einstein's photoelectric equation: $e V_0 = h 
u - h 
u_0$,where $V_0$ is the stopping potential,$
u$ is the frequency of incident

Vedclass · Accepted Answer

$9 	imes 10^{14}$

Vedclass · Answer

(C) According to Einstein's photoelectric equation: $e V_0 = h 
u - h 
u_0$,where $V_0$ is the stopping potential,$
u$ is the frequency of incident light,and $
u_0$ is the threshold frequency.Rearranging for the threshold frequency $
u_0$: $
u_0 = 
u - \frac{e V_0}{h}$.Since $
u = \frac{c}{\lambda}$,we have $
u_0 = \frac{c}{\lambda} - \frac{e V_0}{h}$.Substituting the given values: $c = 3 	imes 10^8 \,m/s$,$\lambda = 2 	imes 10^{-7} \,m$,$e = 1.6 	imes 10^{-19} \,C$,$V_0 = 2.5 \,V$,and $h = 6.6 	imes 10^{-34} \,J-s$.$
u_0 = \frac{3 	imes 10^8}{2 	imes 10^{-7}} - \frac{1.6 	imes 10^{-19} 	imes 2.5}{6.6 	imes 10^{-34}}$$
u_0 = 1.5 	imes 10^{15} - 0.606 	imes 10^{15} \approx 0.894 	imes 10^{15} \,Hz$.Rounding to the nearest significant value,$
u_0 \approx 9.0 	imes 10^{14} \,Hz$.

Similar Questions

Light of incident frequency $2$ times the threshold frequency is incident on a photosensitive material. If the incident frequency is made $\left(\frac{1}{3}\right)^{\text{rd}}$ and intensity is doubled,then the photoelectric current will

$A$ light whose frequency is equal to $6 \times 10^{14} \, Hz$ is incident on a metal whose work function is $2 \, eV$. $[h = 6.63 \times 10^{-34} \, Js, 1 \, eV = 1.6 \times 10^{-19} \, J]$. The maximum kinetic energy of the emitted electrons will be ............ $eV$. (in $.49$)

Radiation of two photon energies,twice and five times the work function of a metal,are incident successively on the metal surface. The ratio of the maximum velocity of photoelectrons emitted in the two cases is:

Two incident radiations having energies two times and ten times of the work function of a metal surface,produce photoelectric effect. The ratio of maximum velocities of emitted photoelectrons respectively is

When the light source is kept $20 \ cm$ away from a photo cell,a stopping potential of $0.6 \ V$ is obtained. When the source is kept $40 \ cm$ away,the stopping potential will be .......... $V$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module