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

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(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 $V_s = \frac{h}{e} \left( \frac{15 \omega}{2 \pi} ight) - \frac{\phi}{e}$.

List-$I$	List-$II$
$(P)$ $2000 \, K$	$(1)$ The radiation at peak wavelength can lead to emission of photoelectrons from a metal of work function $4 \, eV$
$(Q)$ $3000 \, K$	$(2)$ The radiation at peak wavelength is visible to human eye.
$(R)$ $5000 \, K$	$(3)$ The radiation at peak emission wavelength will result in the widest central maximum of a single slit diffraction.
$(S)$ $10000 \, K$	$(4)$ The power emitted per unit area is $1/16$ of that emitted by a blackbody at temperature $6000 \, K$.
	$(5)$ The radiation at peak emission wavelength can be used to image human bones.

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:

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