Exams › JEE Main › Physics

A He⁺ ion is in its first excited state. Its ionization energy is:

1. 48.36 eV
2. 13.60 eV
3. 54.40 eV
4. 6.04 eV

Correct answer: 13.60 eV

Solution

The ionization energy of a He⁺ ion in its first excited state corresponds to the energy required to remove an electron from that state, which is calculated using the formula for hydrogen-like atoms. For He⁺, the ionization energy from the ground state is 54.4 eV, and the first excited state corresponds to 13.6 eV, making option B the correct answer.

Related JEE Main Physics questions

• An electron in one Bohr orbit with principal quantum number n = n1 can produce 3 distinct spectral lines on de-excitation. In another orbit with principal quantum number n = n2, it can produce 6 distinct spectral lines. What is the ratio of the orbital speeds of the electron in these two orbits?
• An electron in a hydrogen atom is found in an excited state with energy equal to -3.4 eV. What is its angular momentum?
• A beam of electrons having energy 12.5 eV is directed at hydrogen gas in the gaseous state at room temperature. The emitted spectrum will contain:
• A hydrogen atom and a Li++ ion are each in their second excited state. If their electronic angular momenta are denoted by L_H and L_Li, and their energies by E_H and E_Li respectively, which statement is correct?
• For a proton–electron system, the interaction potential is V = V0 ln(r/r0), where r0 is a constant. If Bohr’s model is assumed to hold, how does the radius rₙ depend on the principal quantum number n?
• In the hydrogen spectrum, the longest wavelength in the ultraviolet region is 122 nm. The shortest wavelength in the infrared region, rounded to the nearest whole number, is

⚔️ Practice JEE Main Physics free + battle 1v1 →