A mixture of hydrogen atoms (in their ground state) and hydrogen like ions (in their first excited state) are being excited by electrons which have been accelerated by same potential of emitted light are found in the ratio 5:1. Then, find
(a) the minimum value of V for which both the atoms get excited after collision with electrons.
(b) atomic number of other ion.
(c ) the energy of emitted light.

Answer

A, B, C, D

Explanation

(a) and (b) when hydrogen atom is excited, then
`eV = E_0 ((1)/(1) - (1)/(n^2)) ….(i)`
When ion is excited,
`eV = E_0Z^2[(1)/(2^2) - (1)/(n_1^2)] …..(ii)`
Wavelength of emitted light,
`(hc)/(lambda_1) = E_0((1)/(1) - (1)/(n^2)) ......(iii)`
`(hc)/(lambda_2) = E_0Z^2 ((1)/(1) - (1)/(n_1^2)) .....(iv)`
Further it is given that
`(lambda_1)/(lambda_2) = (5)/(1) ....(v)`
Solving the above equations, we get
`Z =2, n=2, n_1 =4`
and V = 10.2 V
(c) Energy of emitted photon by the hydrogen
atom `= E_2 - E_1`
= 10.2 eV
and by the ion `= E_4 - E_1`
`=(13.6)(2)^2 ((1-(1)/(16))`
= 51eV.