The wave equation of an electric field at a p | vedclass

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

Question

(B) The electric field is given by $E = 100 [\sin(7 \omega t) + \cos(10 \omega t) + \cos(15 \omega t)]$.The frequencies present in the wave are $\omeg

Vedclass · Accepted Answer

$\frac{h}{e} \left( \frac{15 \omega}{2 \pi} \right) - \frac{\phi}{e}$

Vedclass · Answer

(B) The electric field is given by $E = 100 [\sin(7 \omega t) + \cos(10 \omega t) + \cos(15 \omega t)]$.The frequencies present in the wave are $\omega_1 = 7 \omega$,$\omega_2 = 10 \omega$,and $\omega_3 = 15 \omega$.The angular frequencies are related to linear frequencies by $\omega = 2 \pi 
u$,so $
u = \frac{\omega}{2 \pi}$.The frequencies are $
u_1 = \frac{7 \omega}{2 \pi}$,$
u_2 = \frac{10 \omega}{2 \pi}$,and $
u_3 = \frac{15 \omega}{2 \pi}$.The maximum frequency is $
u_{max} = \frac{15 \omega}{2 \pi}$.According to Einstein's photoelectric equation,the maximum kinetic energy is $K_{max} = h 
u_{max} - \phi$.The stopping potential $V_s$ is given by $K_{max} = e V_s$,so $V_s = \frac{K_{max}}{e} = \frac{h 
u_{max} - \phi}{e}$.Substituting $
u_{max}$,we get $V_s = \frac{h}{e} \left( \frac{15 \omega}{2 \pi} ight) - \frac{\phi}{e}$.

The wave equation of an electric field at a point is $E = 100 \frac{V}{m} [\sin(7 \omega t) + \cos(10 \omega t) + \cos(15 \omega t)]$ at instant $t$. If the work function of the photocell is $\phi$,then the stopping potential is:

Similar Questions

The maximum kinetic energy of emitted electrons in a photoelectric effect does not depend upon

$A$ silver ball is suspended by a string in a vacuum chamber and ultraviolet light of wavelength $200 \ nm$ is incident on it. The maximum potential acquired by the ball,if the work function of silver is $4.7 \ eV$,will be .............. $V$.

In an experiment of photoelectric effect,the stopping potential was measured to be $V_{1}$ and $V_{2}$ with incident light of wavelength $\lambda$ and $\frac{\lambda}{2}$ respectively. The relation between $V_{1}$ and $V_{2}$ is:

Light of two different frequencies whose photons have energies $1 \text{ eV}$ and $2.5 \text{ eV}$ respectively,successively illuminate a metallic surface whose work function is $0.5 \text{ eV}$. The ratio of the maximum speeds of the emitted electrons will be:

The work function of a metal is $4.2 \ eV$. Its threshold wavelength will be .......... $ \mathring A $.

Vedclass Test Series

Exam Paper Generator

Online Exam Module