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Question

(D) The magnetic field is given by $B = B_0 [\sin(\omega_1 t) + \sin(\omega_2 t)]$,where $\omega_1 = 3.14 	imes 10^7 c$ and $\omega_2 = 6.28 	imes 1

Vedclass · Accepted Answer

$7.72$

Vedclass · Answer

(D) The magnetic field is given by $B = B_0 [\sin(\omega_1 t) + \sin(\omega_2 t)]$,where $\omega_1 = 3.14 	imes 10^7 c$ and $\omega_2 = 6.28 	imes 10^7 c$.Since $\omega = 2\pi 
u$,the frequencies are $
u_1 = \frac{3.14 	imes 10^7 c}{2\pi} = 0.5 	imes 10^7 c$ and $
u_2 = \frac{6.28 	imes 10^7 c}{2\pi} = 1.0 	imes 10^7 c$.The maximum frequency is $
u_{\max} = 1.0 	imes 10^7 	imes (3 	imes 10^8) = 3 	imes 10^{15} \ Hz$.The energy of the photon is $E = h
u_{\max} = (6.63 	imes 10^{-34} \ J\cdot s) 	imes (3 	imes 10^{15} \ Hz) = 1.989 	imes 10^{-18} \ J$.Converting to $eV$: $E = \frac{1.989 	imes 10^{-18}}{1.6 	imes 10^{-19}} \approx 12.43 \ eV$.The maximum kinetic energy is $KE_{\max} = E - \phi = 12.43 \ eV - 4.7 \ eV = 7.73 \ eV$. The closest option is $7.72 \ eV$.

The magnetic field associated with a light wave is given,at the origin,by $B = B_0 [\sin(3.14 \times 10^7 ct) + \sin(6.28 \times 10^7 ct)]$. If this light falls on a silver plate having a work function of $4.7 \ eV$,what will be the maximum kinetic energy of the photoelectrons? (in $eV$)

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