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 photoelectric threshold wavelength for a certain metal surface is $3300 \,\mathring{A}$. What is the maximum kinetic energy of the photoelectrons emitted if radiations of wavelength $1100 \,\mathring{A}$ are used?

In a photoelectric experiment, the wavelength of the light incident on the metal is changed from $200 \, nm$ to $400 \, nm$. The decrease in the stopping potential is close to [Use $hc = 1240 \, eV \cdot nm$ where $h$ is Planck's constant and $c$ is the velocity of light]. (in $ \, V$)

The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy $6 eV$ fall on it is $4 eV$. The stopping potential in volts is

In the photoelectric effect,a graph is plotted between the maximum kinetic energy of electrons and the frequency of incident photons. The slope of this curve will be .......

Vedclass Test Series

Exam Paper Generator

Online Exam Module