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NEET-UG Biology

Matter waves and the de Broglie relation — practice questions

83 questions in the bank on this idea. Below are 10 of them, exactly as they appear in a test.

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  1. Question 1 · difficulty L1 · recall

    The de Broglie wavelength of a particle with momentum p is

    • A. λ=hp
    • B. λ=p/h
    • C. λ=h/p
    • D. λ=hc/p
  2. Question 2 · difficulty L1 · understanding

    For a nonrelativistic particle of mass m and speed v, de Broglie wavelength is

    • A. hmv
    • B. mv/h
    • C. hv/m
    • D. h/(mv)
  3. Question 3 · difficulty L2 · understanding

    Electron diffraction provides evidence for

    • A. the wave nature of matter
    • B. only classical particle trajectories
    • C. absence of electron momentum
    • D. magnetic monopoles
  4. Question 4 · difficulty L2 · application

    If a particle's momentum doubles, its de Broglie wavelength

    • A. doubles
    • B. halves
    • C. quadruples
    • D. stays unchanged
  5. Question 5 · difficulty L2 · understanding

    An electron is travelling with a velocity vv in free space and when it enters a medium, its velocity is reduced by 20%20 \%. The de Broglie wavelength of electron in the medium is αλ0\alpha \lambda_0, where λ0\lambda_0 is its de Broglie wavelength in free space. The value of α\alpha is ____\_\_\_\_ .

    • A. 1.20
    • B. 1.0
    • C. 1.25
    • D. 0.75
  6. Question 6 · difficulty L2 · understanding

    The de Broglie wavelength associated with an electron accelerated through a potential difference V is λe\lambda_{\mathrm{e}} and the de Broglie wavelength associated with a proton accelerated through the same potential difference is λp\lambda_{\mathrm{p}}. If their corresponding masses are mem_{\mathrm{e}} and mpm_{\mathrm{p}}, respectively, then the ratio of their de Broglie wavelengths (λeλp)\left(\frac{\lambda_e}{\lambda_p}\right) is ____\_\_\_\_ .

    • A.  mpme \text { } \sqrt{\frac{m_p}{m_e}}
    • B. memp \sqrt{\frac{m_e}{m_p}}
    • C. mpme \frac{m_p}{m_e}
    • D. (mpme)2 \left(\frac{m_p}{m_e}\right)^2
  7. Question 7 · difficulty L2 · understanding

    An electron with mass ' m ' with an initial velocity (t=0)v=v0i^(v0>0)(\mathrm{t}=0) \overrightarrow{\mathrm{v}}=\mathrm{v}_0 \hat{i}\left(\mathrm{v}_0>0\right) enters a magnetic field B=B0j^\overrightarrow{\mathrm{B}}=\mathrm{B}_0 \hat{j}. If the initial de-Broglie wavelength at t=0\mathrm{t}=0 is λ0\lambda_0 then its value after time ' t ' would be :

    • A. λ01e2 B02t2 m2\frac{\lambda_0}{\sqrt{1-\frac{\mathrm{e}^2 \mathrm{~B}_0^2 \mathrm{t}^2}{\mathrm{~m}^2}}}
    • B. λ0\lambda_0
    • C. λ01+e2 B02t2 m2\lambda_0 \sqrt{1+\frac{\mathrm{e}^2 \mathrm{~B}_0^2 \mathrm{t}^2}{\mathrm{~m}^2}}
    • D. λ01+e2 B02t2 m2\frac{\lambda_0}{\sqrt{1+\frac{\mathrm{e}^2 \mathrm{~B}_0^2 \mathrm{t}^2}{\mathrm{~m}^2}}}
  8. Question 8 · difficulty L2 · understanding

    A sub-atomic particle of mass 1030 kg10^{-30} \mathrm{~kg} is moving with a velocity 2.21×106 m/s2.21 \times 10^6 \mathrm{~m} / \mathrm{s}. Under the matter wave consideration, the particle will behave closely like \qquad (h=6.63×1034 J.s)\left(\mathrm{h}=6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}\right)

    • A. X-rays
    • B. Infra-red radiation
    • C. Gamma rays
    • D. Visible radiation
  9. Question 9 · difficulty L2 · understanding

    A proton and an electron have the same de Broglie wavelength. If Kp\mathrm{K}_{\mathrm{p}} and Ke\mathrm{K}_{\mathrm{e}} be the kinetic energies of proton and electron respectively, then choose the correct relation :

    • A. Kp>Ke\mathrm{K_p>K_e}
    • B. Kp=Ke\mathrm{K_p=K_e}
    • C. Kp<Ke\mathrm{K}_{\mathrm{p}}<\mathrm{K}_{\mathrm{e}}
    • D. Kp=Ke2\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{e}}{ }^2
  10. Question 10 · difficulty L2 · understanding

    A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is: (Assume h = 6.63 ×1034 J s, me=9.0×1031 kg\times 10^{-34} \mathrm{~J} \mathrm{~s}, \mathrm{~m}_{\mathrm{e}}=9.0 \times 10^{-31} \mathrm{~kg} and mp=1836\mathrm{m}_{\mathrm{p}}=1836 times me\mathrm{m}_{\mathrm{e}} )

    • A. 1:118361: \frac{1}{1836}
    • B. 1:18361: \sqrt{1836}
    • C. 1:18361: 1836
    • D. 1:118361: \frac{1}{\sqrt{1836}}

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