A free electron at rest is captured by a proton and goes directly into the \(n=2\) level of a hydrogen atom. What is the wavelength of the photon emitted in this process? (Take \(E_n=-13.6/n^2\ \mathrm{eV}\) and \(hc=1240\ \mathrm{eV\,nm}\).)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(365\ \mathrm{nm}\)
A free electron at rest has zero total energy. This is the energy of the level \( n=\infty \). When it is captured into the \( n=2 \) level, the energy lost comes out as one photon.
Given:
\( E_i=0 \)
\( E_f=E_2 \)
\( \begin{aligned} E_2 &= -\frac{13.6}{2^2} \\ &= -3.4\ \mathrm{eV} \end{aligned} \)
Energy of the photon:
\( \begin{aligned} E &= E_i-E_f \\ &= 0-(-3.4) \\ &= 3.4\ \mathrm{eV} \end{aligned} \)
Wavelength of the photon:
\( \begin{aligned} \lambda &= \frac{hc}{E} \\ &= \frac{1240}{3.4} \\ &= 365\ \mathrm{nm} \end{aligned} \)
Therefore, the photon emitted when the electron settles in the \( n=2 \) level has a wavelength of about \( 365\ \mathrm{nm} \).
NCERT (2026-27):
Energy levels – “… In this diagram, the highest energy state corresponds to n =∞ in Eq, (12.10) and has an energy of 0 eV. This is the energy of the atom when the electron is completely removed (r = ∞) from the nucleus and is at rest.”
Energy levels – “… From Eq. (12.10), for n = 2; the energy E2 is –3.40 eV.”
\( E_n=-\frac{13.6\ \mathrm{eV}}{n^{2}} \qquad (12.10) \)