de Broglie Waves — MH-CET Physics MCQs with Solutions
Free MH-CET Physics de Broglie Waves MCQs with step-by-step solutions (21 questions). Part of Dual Nature of Radiation and Matter. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — de Broglie Waves · easy · theory
The de Broglie wavelength of a particle of momentum $p$ is:
A. $hp$
B. $\dfrac{p}{h}$
C. $\dfrac{h}{p^2}$
D. $\dfrac{h}{p}$ ✓ Correct
Solution: Equivalently $\lambda = \dfrac{h}{mv}$ — the same relation that holds for a photon.
Q2 — de Broglie Waves · easy · theory
The de Broglie hypothesis states that:
A. Only electrons behave as waves
B. Waves cannot carry momentum
C. Only photons behave as particles
D. All moving matter has an associated wave ✓ Correct
Solution: Proposed in 1924, it extended wave-particle duality from radiation to matter.
Q3 — de Broglie Waves · medium · theory
For particles moving with the same speed, the de Broglie wavelength is:
A. The same for all particles
B. Larger for the lighter particle ✓ Correct
C. Larger for the heavier particle
D. Independent of mass
Solution: $\lambda = \dfrac{h}{mv} \propto \dfrac{1}{m}$ at a given speed.
Q4 — de Broglie Waves · medium · theory
The de Broglie wavelength of a particle of mass $m$ and kinetic energy $K$ is:
A. $\dfrac{h}{2mK}$
B. $\dfrac{h}{\sqrt{2mK}}$ ✓ Correct
C. $\dfrac{h}{\sqrt{mK}}$
D. $\dfrac{\sqrt{2mK}}{h}$
Solution: Using $p = \sqrt{2mK}$ in $\lambda = \dfrac{h}{p}$.
Q5 — de Broglie Waves · medium · theory
The de Broglie wavelength of an electron accelerated through a potential difference $V$ volt is approximately:
A. $12.27\sqrt{V}\text{ \AA}$
B. $\dfrac{12.27}{V}\text{ \AA}$
C. $\dfrac{12.27}{\sqrt{V}}\text{ \AA}$ ✓ Correct
D. $\dfrac{1.227}{\sqrt{V}}\text{ \AA}$
Solution: This convenient formula follows from $\lambda = \dfrac{h}{\sqrt{2meV}}$.
Q6 — de Broglie Waves · medium · theory
The wave nature of macroscopic objects is not observed because:
A. They move too slowly
B. They are electrically neutral
C. They carry no momentum
D. Their de Broglie wavelengths are far too small to detect ✓ Correct
Solution: With a large mass, $\lambda = \dfrac{h}{mv}$ becomes vanishingly small compared with any aperture.
Q7 — de Broglie Waves · easy · theory
The de Broglie wavelength of a particle is:
A. Proportional to the square of its momentum
B. Directly proportional to its momentum
C. Inversely proportional to its momentum ✓ Correct
D. Independent of its momentum
Solution: This is the content of $\lambda = \dfrac{h}{p}$.
Q8 — de Broglie Waves · medium · theory
Wave-particle duality means that:
A. Radiation is a wave and matter is a particle
B. Both radiation and matter exhibit wave and particle properties ✓ Correct
C. Neither waves nor particles really exist
D. Matter is a wave and radiation is a particle
Solution: Which aspect appears depends on the experiment performed, never both at once in the same measurement.
Q9 — de Broglie Waves · medium · theory
The concept of matter waves was experimentally confirmed by:
A. Young's double slit experiment with light
B. Rutherford's scattering experiment
C. The Davisson-Germer electron diffraction experiment ✓ Correct
D. Millikan's oil drop experiment
Solution: Electrons scattered from a nickel crystal produced a diffraction pattern, which only waves can do.
Q10 — de Broglie Waves · easy · numerical
If the momentum of a particle is doubled, its de Broglie wavelength becomes:
A. Four times as large
B. Twice as large
C. Half as large ✓ Correct
D. Unchanged
Solution: $\lambda = \dfrac{h}{p} \propto \dfrac{1}{p}$.
Q11 — de Broglie Waves · medium · numerical
If the kinetic energy of a free electron is increased sixteen times, its de Broglie wavelength becomes:
A. Four times as large
B. Sixteen times as large
C. One-fourth as large ✓ Correct
D. One-sixteenth as large
Solution: $\lambda \propto \dfrac{1}{\sqrt{K}}$, so a sixteen-fold energy divides the wavelength by four.
Q12 — de Broglie Waves · medium · numerical
The de Broglie wavelength of an electron accelerated through $100\text{ V}$ is approximately:
A. $0.012\text{ \AA}$
B. $1.227\text{ \AA}$ ✓ Correct
C. $12.27\text{ \AA}$
D. $0.123\text{ \AA}$
Solution: $\lambda = \dfrac{12.27}{\sqrt{100}} = 1.227\text{ \AA}$.
Q13 — de Broglie Waves · medium · numerical
The de Broglie wavelength of an electron accelerated through $400\text{ V}$ is approximately:
A. $2.45\text{ \AA}$
B. $0.61\text{ \AA}$ ✓ Correct
C. $0.031\text{ \AA}$
D. $1.227\text{ \AA}$
Solution: $\lambda = \dfrac{12.27}{\sqrt{400}} = \dfrac{12.27}{20} \approx 0.61\text{ \AA}$.
Q14 — de Broglie Waves · hard · numerical
An electron, a proton and an alpha particle have equal kinetic energies. Their de Broglie wavelengths satisfy:
A. $\lambda_p > \lambda_e > \lambda_\alpha$
B. $\lambda_e > \lambda_p > \lambda_\alpha$ ✓ Correct
C. $\lambda_\alpha > \lambda_p > \lambda_e$
D. $\lambda_e = \lambda_p = \lambda_\alpha$
Solution: At equal $K$, $\lambda \propto \dfrac{1}{\sqrt{m}}$, and the electron is by far the lightest.
Q15 — de Broglie Waves · hard · numerical
The de Broglie wavelength of an electron moving at $10^6\text{ m/s}$ is approximately ($m_e = 9.1 \times 10^{-31}\text{ kg}$):
A. $6.6 \times 10^{-34}\text{ m}$
B. $7.3 \times 10^{-10}\text{ m}$ ✓ Correct
C. $1.4 \times 10^{-9}\text{ m}$
D. $7.3 \times 10^{-12}\text{ m}$
Solution: $\lambda = \dfrac{h}{mv} = \dfrac{6.63 \times 10^{-34}}{9.1 \times 10^{-31} \times 10^6} \approx 7.3 \times 10^{-10}\text{ m}$.
Q16 — de Broglie Waves · hard · numerical
An electron and a photon both have a wavelength of $1.0\text{ nm}$. The ratio of the photon energy to the electron kinetic energy is approximately:
A. $410 : 1$
B. $1 : 1$
C. $1 : 820$
D. $820 : 1$ ✓ Correct
Solution: $\dfrac{E_{photon}}{K_e} = \dfrac{hc/\lambda}{h^2/2m\lambda^2} = \dfrac{2mc\lambda}{h} \approx 820$.
Q17 — de Broglie Waves · medium · numerical
The de Broglie wavelength of an electron accelerated through $25\text{ V}$ is approximately:
A. $12.27\text{ \AA}$
B. $2.45\text{ \AA}$ ✓ Correct
C. $1.227\text{ \AA}$
D. $0.49\text{ \AA}$
Solution: $\lambda = \dfrac{12.27}{\sqrt{25}} = \dfrac{12.27}{5} \approx 2.45\text{ \AA}$.
Q18 — de Broglie Waves · easy · numerical
If the speed of a particle is doubled, its de Broglie wavelength becomes:
A. Twice as large
B. Half as large ✓ Correct
C. Four times as large
D. Unchanged
Solution: $\lambda = \dfrac{h}{mv} \propto \dfrac{1}{v}$.
Q19 — de Broglie Waves · hard · numerical
A proton and an electron move with the same speed. The ratio $\lambda_p : \lambda_e$ is:
A. $m_p : m_e$
B. $\sqrt{m_e} : \sqrt{m_p}$
C. $m_e : m_p$ ✓ Correct
D. $1 : 1$
Solution: At equal speed $\lambda \propto \dfrac{1}{m}$, so the heavier proton has the shorter wavelength.
Q20 — de Broglie Waves · hard · numerical
A cricket ball of mass $0.15\text{ kg}$ moves at $20\text{ m/s}$. Its de Broglie wavelength is approximately:
A. $2.2 \times 10^{-34}\text{ m}$ ✓ Correct
B. $3.0 \times 10^{-33}\text{ m}$
C. $6.6 \times 10^{-34}\text{ m}$
D. $2.2 \times 10^{-30}\text{ m}$
Solution: $\lambda = \dfrac{6.63 \times 10^{-34}}{0.15 \times 20} \approx 2.2 \times 10^{-34}\text{ m}$ — far too small ever to detect.
Q21 — de Broglie Waves · medium · numerical
If the accelerating voltage for an electron beam is quadrupled, the de Broglie wavelength becomes:
A. Twice as large
B. One-fourth as large
C. Half as large ✓ Correct
D. Four times as large
Solution: $\lambda \propto \dfrac{1}{\sqrt{V}}$, so four times the voltage halves the wavelength.