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Photoelectric threshold of silver is lam...

Photoelectric threshold of silver is `lamda=3800A`. Ultraviolet light of `lamda=2600A` is incident of a silver surface.(Mass of the electron `9.11xx10^(-31)kg`)`
2. Calculate the maximum kinetic energy (in eV) of the emitted photoelectrons.

A

1.51

B

2.36

C

3.85

D

4.27

Text Solution

Verified by Experts

The correct Answer is:
A
`lamda_0=3800A`
`W=hf_0=h(c)/(lamda_0)=(6.633xx10^(-34)xx3xx10^(8))/(3800xx10^(-10))`
`=5.23xx10^(-19)J=3.27eV`
Incident wavelength `lamda=2600A`
`f=` incident frequency `=(3xx10^(8))/(2600xx10^(-10))Hz`
Then, `KE_(max)=hf-W-0`
`hf=(6.63xx10^(-34)xx3xx10^(8))/(2600xx10^(-10))`
`=7.65xx10^(-19)J=4.78eV`
`KE_(max)=hf-W-0=4.78eV-3.27eV=1.51eV`
`KE_(max)=(1)/(2)mv_(max)^(2)`
or `v_(max)=sqrt((2KE_(max))/(m))`
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