# The magnetic field associated with a light wave is given, at the origin, by B=B0[sin(3.14 107)ct + sin(6.28 7ct).  If this light gals on a silver plate having a work function of 4.7 eV, What will be the maximum kinetic energy of the photo electrons? (c=3 108ms−1,h=6.6 10−34 J-s) a. 7.72eV b. 8.52eV c. 12.5eV d. 6.82eV

## Question ID - 50156 :- The magnetic field associated with a light wave is given, at the origin, by B=B0[sin(3.14 107)ct + sin(6.28 7ct).  If this light gals on a silver plate having a work function of 4.7 eV, What will be the maximum kinetic energy of the photo electrons? (c=3 108ms−1,h=6.6 10−34 J-s) a. 7.72eV b. 8.52eV c. 12.5eV d. 6.82eV

3537

B1=B0sin( 107C)t + B0 sin(2 10−7C)t since there are two EM waves with different frequency, to get maximum kinetic energy we take the photon with higher frequency

B1=B0sin( 107C)t  V1=

B2=B0sin(2 107C)t  V2= 10C

Where C is speed of light C=3 10m/s

V2 V1

So KE of photoelectron will be maximum for photon of higher energy.

V2 =107CHz

hv= +KEmax

energy of photon

Eph=hv=6.6 10−34 10−7 3 109

Eph=6.6 3 10−19J

=

KEmax =Eph

=12.375−4.7

=7.675eV  7.7 eV

Next Question :
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