The magnetic field associated with a light wave is given, at the origin, by B=B_{0}[sin(3.14 10^{7})ct + sin(6.28 ^{7}ct). 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 10^{8}ms^{−1},h=6.6 10^{−34} J-s)

a. |
7.72eV |
b. |
8.52eV |

c. |
12.5eV |
d. |
6.82eV |

The magnetic field associated with a light wave is given, at the origin, by B=B_{0}[sin(3.14 10^{7})ct + sin(6.28 ^{7}ct). 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 10^{8}ms^{−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

B_{1}=B_{0}sin( 10^{7}C)t + B_{0} sin(2 10^{−7}C)t since there are two EM waves with different frequency, to get maximum kinetic energy we take the photon with higher frequency

B_{1}=B_{0}sin( 10^{7}C)t V_{1}=

B_{2}=B_{0}sin(2 10^{7}C)t V_{2}= 10^{7}C

Where C is speed of light C=3 10^{8 }m/s

V_{2} V_{1}

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

V^{2} =10^{7}CHz

hv= +KE_{max }

energy of photon

E_{ph}=hv=6.6 10^{−34} 10^{−7} 3 10^{9}

E_{ph}=6.6 3 10^{−19}J

=

KE_{max} =E_{ph} –

=12.375−4.7

=7.675eV 7.7 eV

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 cm |
(b) |
22.75 cm |
(c) |
45 cm |
(d) |
45.5 cm |

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