The photoelectric threshold wavelength of silver is `3250 xx 10^(-10) m`. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength `2536 xx 10^(-10) m` is
`(Given h = 4.14 xx 10^(6) ms^(-1) eVs` and `c = 3 xx 10^(8) ms^(-1))`

Options

A. `6xx10^5 ms^(-1)`
B. `6xx10^6ms^(-1)`
C. `61xx10^3ms^(-1)`
D. `0.3xx10^5ms^(-1)`

Answer

A

Explanation

Threshold wavelength,
`lamda_0=3250 xx10^(-10) m = 325nm, `
Wavelength of the incident light,
`lamda=2536xx10^(-10) m = 253.6 nm`
The work function for the silver surface is given by
`phi=(hc)/lamda_0-(1242eV-nm)/(325 nm)-3.82 eV`
Energy of the incident photon
`(hc)/lamda=(1242 eV 0nm)/(253.6nm)=4,89 eV`
Maximum kinetic energy of the ejected electron can be calculated as follows :
`K.E._(max)=(4.89-3.82) eV=1.077 eV`
Let the maximum velocity of the ejected electron by v.
Thus,
`1/2mv^2=1.077xx1.6xx10^(-19)`
`v=sqrt((2xx1.077xx1.6xx10^(-19))/(9.1xx10^(-31)))`
`v = 6 xx 10 ^5 ms ^(-1)`