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

Answer

B

Explanation

An electron accelerated through a P.D. of V volts, gets the K.E., E=eV. Due to , it is lifted to the first excited state and when it returns to the normal state from `n_2 " to " n_1` the energy eV is emitted as a radiation of wavelength `lamda`
Thus eV = E `=(hc)/(lamda)`
`therefore lamda = (hc)/(eV) = (6.6xx10^(-34) xx3xx10^8)/(1.6xx10^(-19) xx4.9) m`
`= (19.8)/(4.9xx1.6)xx10^(-34+8+9)`
`=2.525xx10^(-7) m = 2525 Å`