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Chapter 6: Q9E (page 1275)

When ultraviolet light with a wavelength of 4.00nm falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be 1.10eV. What is the maximum kinetic energy of the photoelectrons when light of wavelength 300.0nm falls on the same surface?

Short Answer

Expert verified

The maximum kinetic energy of the photoelectrons when light of wavelength 300.0nm falls on the surface is 2.14eV.

Step by step solution

01

Formula for maximum kinetic energy and relation between f,c and λ

K122maxmax (1)

Where h=4.136*10-15eVs(Planck鈥檚 constant)

f=c位

02

Calculate the work function

First calculate the work function atKmaxand位=400nm

So, for that calculate the f,

f=c位

Here,

鈬�位=400109=4*10-7m

So,

f=3*108m/s4*10-7m=7.5*1014Hz

Substitute the values in equation (1),

1.10eV=(4.136*10-15eVs)(7.5*1014s-1)-蠒鈬�蠒=2.00eV

03

Calculate the maximum kinetic energy

For位=300nm,

位=3*10-7m

So,

f2=3*108m/s3*10-7m)=1*1015Hz

Therefore,

K-15152max

Thus, the maximum kinetic energy of the photoelectrons when light of wavelength falls on the surface is 2.14eV.

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