Exams › JEE Advanced › Physics

A hydrogen atom has its electron in the nth orbital. Electromagnetic radiation of wavelength 90 nm is incident on it and ionises the atom. The kinetic energy of the ejected electron is 10.4 eV. Find the value of n. (Use hc = 1242 eV nm)

1. 1
2. 2
3. 3
4. 4

Correct answer: 2

Solution

The photon supplies energy hc/lambda = 1242/90 = 13.8 eV. This energy must ionise the electron (overcoming its binding energy |Eₙ|) and give it kinetic energy 10.4 eV. So |Eₙ| = 13.8 - 10.4 = 3.4 eV. Using Eₙ = -13.6/n²: 3.4 = 13.6/n² -> n² = 4 -> n = 2.

Related JEE Advanced Physics questions

• What is the kinetic energy of an electron in the nth orbit?
• The wavelengths of the Kα X-ray lines for copper (atomic number 29) and molybdenum (atomic number 42) are denoted as λCu and λMo, respectively. What is the approximate value of the ratio λCu to λMo?
• For large quantum numbers (n >> 1), what is the dependence of the frequency of the photon emitted when an electron in a hydrogen atom transitions from the nth orbit to the (n-1)th orbit?
• A hydrogen atom initially in its 6th excited state (principal quantum number n = 7) de-excites to the ground state through one or more electronic transitions. Given that no spectral lines are produced in either the Paschen series (n_final = 3) or the Brackett series (n_final = 4), what is the maximum number of spectral lines of distinct photon energies that can be observed?
• In a hydrogen atom, the orbital radius of an electron changes from 2.12 Angstrom to 4.77 Angstrom. Which of the following correctly describes the energy transition that occurred?
• What is the potential energy of the electron in the 2nd Bohr orbit of Li²+? (Given r₀ = radius of the 1st Bohr orbit of hydrogen)

⚔️ Practice JEE Advanced Physics free + battle 1v1 →