Question

The photoelectric work function of potassium is 2.3eV. If light with wavelength 250nm falls on potassium find (a) the stopping potential in volts. (b) the kinetic energy in eV of the most energetic electron ejected (c)the speed of these electrons (d)If potassium is replaced with Aluminium, which has a greater work function (4.3eV), compute the change in the threshold frequency.

Answer 1

a)

we know that

KE = hf - W

also

frequency = speed of light (c) / wave length (lamda )

so

KE = hc/lamda - W

now

h = 4.136 x 10-15 eV . s

c = 3 x 10^8 m/s

lamda = 250 x 10-9 m

work function (W) = 2.3 eV

so

KE = ( 4.136 x 10-15 x 3 x 10^8 / 250 x 10-9 ) - 2.3

KE = 2.6632

we know that

kinetic energy KE = eVo

so

eVo = 2.6632 eV

so

Vo = 2.6632

so the stopping potential is 2.6632 Volts

b)

now

K.E max = stopping potenial = 2.6632 eV

c)

now

K.E Max = 2.6632 eV

0.5 mV2 = 2.6632 x 1.6 x 10-19 J

mass of electron = 9.11 x 10-31 kg

so

0.5 x 9.11 x 10-31 x v2 = 2.6632 x 1.6 x 10-19

v2= 9.35 x 10^11

v= sqrt ( 9.35 x 10^11)

v =9.67 x 10^5

so

the speed of those electrons is 9.67 x 10^5 m/s

d)

we know that

work function (W) = hfo

fo is threshold frequency

h = 4.136 x 10-15 eV . s

so

for aluminium

4.3 = 4.136 x 10-15 x fo

fo= 1.04 x 10^15 s-1

For potassium

2.3 - 4.136 x 10-15 x fo

fo = 5.56 x 10^14

so the change in treshold frequency = ( 1.04 x 10^15 - 5.56 x 10^14)

change in threshold frequency = 4.84 x 10^14