An electron passing through a potential difference of `4.9 V` collides with a mercury atom and transfers it to the first excited state. What is the wavelength of a photon corresponding to the transition of the mercury atom to its normal state?

Options

A. `2050 Å`
B. `2240 Å`
C. `2525 Å`
D. `2935 Å`

Answer

C

Explanation

`(hc)/(lambda) = E = eV`
`rArr lambda = (hc)/(eV) - (6.6 xx 10^(-34) xx 3xx 10^(8))/(1.6 xx 10^(-19) xx 4.9) = 2525 Å`