If radiation corresponding to first line of "Balmer series" of $$H{e^ + }$$ ion knocked out electron from 1st excited state of $$H$$ atom, the kinetic energy of ejected electron from $$H$$ atom would be $$\left( {eV} \right) - $$
[Given $${E_n} = - \frac{{{Z^2}}}{{{n^2}}}\left( {13.6\,eV} \right)$$ ]
A.
$$4.155\,eV$$
B.
$$8.310\,eV$$
C.
$$2.515\,eV$$
D.
$$5.550\,eV$$
Answer :
$$4.155\,eV$$
Solution :
Energy of photon corresponding to first line of Balmer series $$ = \left( {13.6} \right){\left( 2 \right)^2}\left[ {\frac{1}{4} - \frac{1}{9}} \right]$$
Energy need to eject electron from $$n = 2$$ level in $$H$$ atom $$ = \left( {13.6} \right)\left( {\frac{1}{4}} \right)$$
So, required kinetic energy
$$\eqalign{
& = \left( {13.6} \right)\left[ {\left( {\frac{1}{4} - \frac{1}{9}} \right) - \left( {\frac{1}{4}} \right)} \right]eV \cr
& = 13.6 \times \left( {\frac{{11}}{{36}}} \right) \cr
& = 4.155\,eV \cr} $$
Releted MCQ Question on Modern Physics >> Atoms And Nuclei
Releted Question 1
If elements with principal quantum number $$n > 4$$ were not allowed in nature, the number of possible elements would be
An energy of $$24.6\,eV$$ is required to remove one of the electrons from a neutral helium atom. The energy in $$\left( {eV} \right)$$ required to remove both the electrons from a neutral helium atom is
As per Bohr model, the minimum energy (in $$eV$$ ) required to remove an electron from the ground state of doubly ionized $$Li$$ atom $$\left( {Z = 3} \right)$$ is