SaraNextGen Study App [2022-23]
Last Updated On 23-10-2022 - By Sara

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

1 Answer
5876 Votes
3537

Answer Key : (a) -

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 :

Let ABCD be a parallelogram and ABEF be a rectangle with EF lying along the line C. If AB = 7 cm and BE = 6.5 cm, then the area of parallelogram is  

(a)

11.375 cm2

(b)

22.75 cm2

(c)

45 cm2

(d)

45.5 cm2


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