The radius of the innermost electron orbit of a hydrogen atom is \(5.3\times10^{-11}\ \mathrm{m}\). In another orbit, the de Broglie wavelength of the electron is \(6.66\times10^{-10}\ \mathrm{m}\). The total energy of the electron in this orbit is (Take \(E_n=-13.6/n^2\ \mathrm{eV}\))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(-3.40\ \mathrm{eV}\)
In the \( n^{\text{th}} \) orbit the radius is \( n^2r_1 \), and \( n \) wavelengths fit into the circumference. So the wavelength in that orbit is \( 2\pi nr_1 \), which gives \( n \).
\( \begin{aligned} \lambda &= \frac{2\pi r_n}{n} \\ &= 2\pi nr_1 \end{aligned} \)
\( \begin{aligned} n &= \frac{\lambda}{2\pi r_1} \\ &= \frac{6.66\times10^{-10}}{2\times3.14\times5.3\times10^{-11}} \\ &= 2 \end{aligned} \)
\( \begin{aligned} E_2 &= -\frac{13.6}{2^2} \\ &= -3.40\ \mathrm{eV} \end{aligned} \)
Hence, the electron is in the \( n=2 \) orbit, with total energy \( -3.40\ \mathrm{eV} \).
NCERT (2026-27):
De Broglie’s Explanation of Bohr’s Second Postulate of Quantisation – “… For an electron moving in nth circular orbit of radius rn, the total distance is the circumference of the orbit, 2πrn. Thus”
\( 2\pi r_n=n\lambda,\ n=1,2,3\ldots \)\( \qquad (12.12) \)
Bohr Model of the Hydrogen Atom – “The radius of nth possible orbit thus found is”
\( \displaystyle r_n=\frac{n^2}{m}\left(\frac{h}{2\pi}\right)^2\frac{4\pi\varepsilon_0}{e^2} \)\( \qquad (12.7) \)
Bohr Model of the Hydrogen Atom – “Atomic energies are often expressed in electron volts (eV) rather than joules. … can be rewritten as”
\( \displaystyle E_n=-\frac{13.6\ \text{eV}}{n^2} \)\( \qquad (12.10) \)