The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and
(b) the threshold frequency of the radiation. If the caesium element is irradiated with a
wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.
(1 eV = 1.602
´ 10–19 J)
Answers
Answered by
1
Explanation:
The work function of cesium atom is 1.9 eV. Converting the unit in J.
1.9eV×1.602×10
−19
J/eV=3.04×10
−19
J
The threshold frequency is
6.626×10
−34
3.04×10
−19
=4.59×10
14
Hz.
The threshold wavelength is λ
0
=
ν
0
c
=
4.59×10
14
3.0×10
8
=6.54×10
−7
m.
The energy of radiated light is
E=
λ
hc
=
500×10
−9
6.626×10
−34
×3×10
8
=3.98×10
−19
J
The kinetic energy of ejected electron is
K.E=3.98×10
−19
−3.04×10
−19
=9.4×10
−20
J=
2
1
mv
2
v=
m
2K.E
=
2
9.1×10
−31
9.14×10
−20
=4.54×10
5
m/s
Similar questions