Question

When a certain metal was irradiated with light of
frequency 1.6x1016 Hz , the photoelectrons emitted
had twice the kinetic energy as the photoelectrons
emitted when the same metal was irradiated with
light of frequency 1.0x1016 Hz. Calculate the
threshold frequency (Vo) for the metal.​

Answer 1

Explanation:

for frequency= 1.6*10^16s-¹

E1= h υ

= h*1.6*10^16s-¹

KE= hυ+hυo

KE1=h*1.6*10^16s-¹+hυo

for frequency= 1.0"10^16s-¹

E2= h*1.0*10^16s-¹

KE2=h*1.0*10^16s-¹+hυo

according to question

KE1=2KE2

h*1.6*10^16s-¹+hυo=2(h*1.0*10^16s-¹+hυo)

h*1.6*10^16s-¹+hυo=2*h*1.0*10^16s-¹+2hυo

h(1.6-1.0)*10^16=hυo

υo=0.6*10^16s-¹

= 6*10^15s-¹

Answer 2

Answer:

E(Total Energy)=Eo(Minimum Energy reqd. for the electron)+K.E(kinetic Energy of Photo electron)

hv=hv0 + 1/2 mv2

6.6*10-34*(3.2 * 1016) =Eo+ K.E

 2.112*10-17 =Eo+ K.E......................(1)

6.6*10-34*(2*1016) =Eo+ 2*K.E.

1.32*10-17=Eo+ 2 K.E  .......................(2)

solving equation (1) and (2): ,,,,,,,,,,,,,,,,,,,,,

Eo=2.904*10-17 j

Explanation:

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