Radiation of wavelength `lamda` is incident on a photocell. The fastest emitted electron has speed v. If the wavelength is charged to `(3lamda)/(4)`, the speed of the fastest emitted electron will be

Options

A. `gt v ((4)/(3))^(1//2)`
B. `lt v ((4)/(3))^(1//2)`
C. `= v ((4)/(3))^(1//2)`
D. `= v ((3)/(4))^(1//2)`

Answer

A

Explanation

In the first case, `(1)/(2) mv^(2) = (hc)/(lamda) - W_(0)`…(1)
In the second case,
`(1)/(2) mv_(1)^(2) = (4hc)/(3lamda) - W_(0)` [where `v_(1)`= speed of fastest emitted electron when wavelength is `(3lamda)/(4)`]
or, `(1)/(2) mv_(1)^(2) = (1)/(2) mv^(2) + (hc)/(3lamda)` [using equation(1)]
or, `v_(1) = sqrt(v^(2) + (2hc)/(3 lamdam)) = sqrt(v^(2) + (2)/(3m) ((1)/(2) mv^(2) + W_(0)))`
`= sqrt((4v^(2))/(3) + (2W_(0))/(3m))`
`:. v_(1) gt sqrt((4)/(3) v^(2)) or, v_(1) gt sqrt((4)/(3))v`