Question

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)​

Answer 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