In a photoemissive cell, with exciting wavelength `lambda`, the faster electron has speed v. If the exciting wavelength is changed to `3lambda//4`, the speed of the fastest electron will be

Options

A. `v((3)/(4))^((1)/(2))`
B. `v((4)/(3))^((1)/(2))`
C. less than `v((4)/(3))^((1)/(2))`
D. greater than `v((4)/(3))^((1)/(2))`.

Answer

B

Explanation

According to Einstein's photoelectric equation,
`(hc)/(lamda_(1))=W_(0)+(1)/(2)mv_(1)^(2) and (hc)/(lamda^(2))=W_(0)+(1)/(2)mv_(2)^(2)`
These expression show that, `v^(2)=((1)/(lamda))`
`therefore(v_(1))/(v_(2))=sqrt((((1)/(lamda_(1))))/(((1)/(lamda_(2)))))=sqrt((lamda_(2))/(lamda_(1)))=sqrt((3lamda//4)/(lamda))`
`thereforev_(2)=v((4)/(3))^((1)/(2))`