Light of wavelength 2000 Å falls on a metal surface of work function 5.2 eV. (a) What is the kinetic energy (in eV) of the fastest electrons emitted from the surface?
Answers
Explanation:
Energy corresponding to incident photon
hv=
λ
hc
=
2000×10
−10
6.6×10
−34
×3×10
8
=9.9×10
−19
J
=
1.6×10
−19
9.9×10
−19
eV=6.2eV
a. The kinetic energy of fastest electrons,
E
k
=hv−W
or E
k
=6.2eV−4.2eV=2eV
b. The kinetic energy of the slowest electron is zero, since the emitted electrons have all possible energies from 0 to certain maximum value E
k
.
c. If V
s
is the stopping potential, then
E
k
=eV
s
or V
S
=
e
E
k
=
e
2eV
=2V
d. If λ
0
is the cut-off wavelength for a aluminium, then
W=(hc/λ
0
)
or λ
0
=(hc/W)
=
4.2×1.6×10
−19
6.6×10
−34
×3×10
8
=3000×10
−10
m=3000
A
o
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SIMILAR QUESTIONS
star-struck
A stopping potential of 0.82 V is required to stop the emission of photoelectrons from the surface of a metal by light of wavelength 4000
A
o
. For light of wavelength 3000
A
o
, the stopping potential is 1.85 V. If the value of Planck's constant is 6.5×10
−X
[1 electrons volt (eV)=1.6×10
−19
J]. Find X?
Hard
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>
The work function of a metallic surface is 5.01 eV. The photoelectrons are emitted when light of wavelength 2000
A
o
falls on it. The potential difference applied to stop the fastest photoelectrons is: [h=4.14×10
−15
eVs]
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