The work function of a metal is 4.2 eV. Its threshold wavelength will be

Options

A. `4000 Å`
B. `3500 Å`
C. `2955 Å`
D. `2500 Å`

Answer

C

Explanation

`f_0=phi_0/h=(4.2 xx1.6 xx10^(-19))/(6.6 xx10^(-34))`
`implies lamda_0=c/(f_(0))=(3xx10^8xx6.6xx10^(-34))/(4.2 xx1.6 xx10^(-19))`
`lamda_0~~2946 xx10^(-10)m`
`lamda_0~~2955 Å`