A particle moves three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is \( 1.813 \times 10^{-4} \). The mass of the particle is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1.67 \times 10^{-27}\,\mathrm{kg} \)
The de Broglie wavelength of a particle is \( \lambda = h/(mv) \). For the particle \( v_p = 3v_e \), so in the ratio the constant \( h \) cancels.
\( \displaystyle \frac{\lambda_p}{\lambda_e} = \frac{m_e v_e}{m_p v_p} = \frac{m_e}{3m_p} \)
\( \displaystyle m_p = \frac{m_e}{3 \times (\lambda_p/\lambda_e)} = \frac{9.11\times10^{-31}\,\mathrm{kg}}{3 \times 1.813\times10^{-4}} \)
\( \displaystyle m_p = \frac{9.11\times10^{-31}}{5.439\times10^{-4}} \approx 1.675\times10^{-27}\,\mathrm{kg} \)
This is about the mass of a proton or neutron, given in Option (1) as \( 1.67 \times 10^{-27}\,\mathrm{kg} \).
Note: This question is from the NCERT Exemplar. NTA sets numerical questions from the Exemplar too, so do not ignore it while practising numericals.
NCERT Exemplar:
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