Last Updated On 01-06-2023 - By Jeesha 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
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 |
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= 107C
Where C is speed of light C=3
108 m/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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