Question

A single electron, orbits around a stationary nucleus of charge z_e, where z is a constant and e is the electronic charge. It requires 47.2 eV to excite the electron from the second Bohr orbit to 3rd Bohr orbit. Find,
(i) The value of z.
(ii) The energy required to excite the electron from the third to the fourth Bohr orbit.
(iii) The wavelength of electromagnetic radiation required to remove the electron from the first Bohr orbit to infinity.
(iv) The kinetic energy, potential energy and angular momentum of the electron in the first Bohr orbit.
(v) The radius of the first Bohr orbit.
(Ionisation energy of hydrogen atom = 13.6 eV. Bohr radius = 5.3 × 10^–11 m, velocity of light = 3 × 10⁸ m/s and Planck's constant = 6.6 × 10^–34 Js)

Solution

Given,
Amount of energy required to excite the electron from second orbit to 3rd orbit = 47.2 eV

(i) Change in energy for a general hydrogen-like atom is,

E_{n_{2}} - E_{n_{1}} = Z^{2} E_{0} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) eV

where E₀ is the ionisation energy of hydrogen atom.
Now,

∆ E = Z^{2} \times 13.6 (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = 47.2

\Rightarrow

Z^{2} \times \frac{13.6}{36} \times 5 = 47.2

\Rightarrow

Z^{2} = \frac{47.2 \times 36}{13.6 \times 5} = 25 Z = 5

(ii) Energy required to excite the electron from third to fourth orbit,

E_{4} - E_{3} = 5^{2} \times 13.6 (\frac{1}{3^{2}} - \frac{1}{4^{2}}) eV

= 25 \times 13.6 \times \frac{7}{144} = 16.53 eV

(iii) Energy required to excite the electron from ground state to infinity,

E_{\infty} - E_{1} = Z^{2} \times 13.6 (\frac{1}{1^{2}} - \frac{1}{\infty}) = 13.6 \times 25 eV

Thus, wavelength of electromagnetic radiation required to remove the electron from first Bohr orbit to infinity is,

λ = \frac{hc}{∆ E} = \frac{(6.6 \times 10^{- 34}) \times 3 \times 10^{8}}{13.6 \times 25 \times 1.6 \times 10^{- 19}} = 0.03640 \times 10^{- 7} = 36.4 \times 10^{- 10} = 36.4 Å

(iv) Kinetic energy of first Bohr orbit is numerically equal to the energy of the orbit.

Therefore,

E_{1} = - Z^{2} E_{0} = - 25 \times 13.6 eV

∴

K . E . = 25 \times 13.6 \times 1.6 \times 10^{- 19} J = 544 \times 10^{- 19} J

Potential energy of electron

= - 2 \times K . E .

= - 2 \times 544 \times 10^{- 19} J = - 1088 \times 10^{- 19} J

Angular momentum of the electron is given by,

L = mvr = \frac{nh}{2 π} = \frac{h}{2 π}

∵ n = 1

for first bohr orbit.

i.e.,

L = \frac{6.6 \times 10^{- 34}}{2 π} = 1.05 \times 10^{- 34} Js

(v) Radius r₁ of the first Bohr orbit is,

r_{n} = \frac{n^{2} r_{0}}{Z} for n = 1 .

That is,

r_{1} = \frac{1^{2} \times 5.3 \times 10^{- 11}}{5} = 1.06 \times 10^{- 11} m .

E_{\infty} - E_{1} = Z^{2} \times 13.6 (\frac{1}{1^{2}} - \frac{1}{\infty}) = 13.6 \times 25 eV

For Daily Free Study Material Join wiredfaculty

Wave Optics

Some More Questions From Wave Optics Chapter

An atom has a nearly continuous mass distribution in a ___ but has a highly non-uniform mass distribution in (Thomson's model/Rutherford's model)

The positively charged part of the atom possesses most of the mass in ___________ (Rutherford's model/both the models).

Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction