The ionisation potential of hydrogen atom is 13.6 volt. In the lowest energy level, this atom is ionised by absorbing a photon of 800 mathringA. The kinetic energy of the released electron will be :

Question

The ionisation potential of hydrogen atom is 13.6 volt. In the lowest energy level, this atom is ionised by absorbing a photon of 800 mathringA. The kinetic energy of the released electron will be : (A) 15.51 eV (B) 2.91 eV (C) 13.6 eV (D) 1.91 eV

Tardigrade Admin · Accepted Answer

The energy of incident photon in joules is

= \frac{h c}{λ} = \frac{6.62 \times 1 0 ^{- 34} \times 3 \times 1 0 ^{- 8}}{800 \times 1 0 ^{- 10}}

= 2.4825 \times 1 0^{- 18} J

The energy of the photon in

e V

is

= \frac{2.4825 \times 1 0 ^{- 18}}{1.6 \times 1 0 ^{- 19}} = 15.51 e V

The minimum energy needed to ionise the atom

= 13.6 e V

Hence, KE of the emitted electron

= 15.51 - 13.6

= 1.91 e V

Q. The ionisation potential of hydrogen atom is 13.6 volt. In the lowest energy level, this atom is ionised by absorbing a photon of 800A˚. The kinetic energy of the released electron will be :

15.51 eV

2.91 eV

13.6 eV

1.91 eV

The energy of incident photon in joules is =λhc​=800×10−106.62×10−34×3×10−8​ =2.4825×10−18J The energy of the photon in eV is =1.6×10−192.4825×10−18​=15.51eV The minimum energy needed to ionise the atom =13.6eV Hence, KE of the emitted electron =15.51−13.6 =1.91eV

Q. The ionisation potential of hydrogen atom is $13.6$ volt. In the lowest energy level, this atom is ionised by absorbing a photon of $800 \overset{˚}{A}$ . The kinetic energy of the released electron will be :