Wave Particle Duality Questions and Answers

Q: A student sets up Young’s double slit experiment with electrons of momentum \(p\) incident normally on the slits of width \(w\) separated by distance \(d\). In order to observe interference fringes on a screen at a distance \(D\) from the slits, which of the following conditions should be satisfied?

The condition is \(\frac{\hbar}{p} \gt \frac{dw}{D}\).The electrons have de Broglie wavelength \(\lambda = h/p\), of order \(\hbar/p\). The interference fringes on the screen are spaced by about \(\lambda D/d\). Each slit has a finite width \(w\), and electrons passing through different parts of one slit produce fringe patterns shifted relative to each other; the fringes stay visible only if their spacing is larger than this spread, that is roughly\[ \begin{aligned} \frac{\lambda D}{d} &\gt w \\ \Rightarrow \frac{\hbar}{p} &\gt \frac{dw}{D} \end{aligned} \]up to a numerical factor. The other options place \(D\) and \(d\) wrongly: \(Dw/d\) would demand a longer wavelength as the screen moves away, which is the reverse of the real dependence, and the forms with \(d^{2}\) do not involve the slit width at all.

Q: Young’s double slit experiment is performed using a beam of C60 (fullerene) molecules, each molecule being made up of 60 carbon atoms. When the slit separation is 50 nm, fringes are formed on a screen kept at a distance of 1 m from the slits. Now, the experiment is repeated with C70 molecules with a slit separation of 92.5 nm. The kinetic energies of both the beams are the same. The position of the 4th bright fringe for C60 will correspond to the \(n^{th}\) bright fringe for C70. What is the value of \(n\) (rounded off to the nearest integer)?

The value is \(n = 8\).The \(n\)th bright fringe lies at \(y = n\lambda D/d\). With equal kinetic energies, \(\lambda = h/\sqrt{2mK}\), so \(\lambda \propto 1/\sqrt{m}\) and \(\lambda_{60}/\lambda_{70} = \sqrt{70/60}\). Equating the positions,\[ \begin{aligned} \frac{4\lambda_{60}}{50} &= \frac{n\lambda_{70}}{92.5} \\ n &= 4 \times \frac{92.5}{50} \times \sqrt{\frac{70}{60}} \\ &= 7.4 \times 1.080 \approx 7.99 \end{aligned} \]which rounds to 8. Ignoring the mass dependence of the wavelength gives 7.4, rounding to 7; using the mass ratio without the square root gives about 8.6. 5 and 6 do not follow from the fringe condition.

Q: A quantum particle of mass \(10^{-20}\) kg is confined within a length of 1 \(nm\) in one-dimension. The minimum uncertainty in the measurement of velocity of the particle, in units of \(\mu m/s\), rounded off to the nearest integer is:[Assume the minimum uncertainty product \(\Delta x \Delta p_{x} \approx \frac{\hbar}{2}\) , use Planck’s constant h = \(6.64 \times 10^{-34} J\ s\)]

The minimum velocity uncertainty is about 5 \(\mu m/s\).With \(\Delta x = 1\) nm and \(\Delta p_x = m\,\Delta v\), the minimum uncertainty product gives\[ \begin{aligned} \Delta v &= \frac{\hbar}{2m\,\Delta x}, \quad \hbar = \frac{6.64 \times 10^{-34}}{2\pi} \\ &= \frac{1.057 \times 10^{-34}}{2 \times 10^{-20} \times 10^{-9}} \\ &= 5.3 \times 10^{-6}\ \text{m/s} \end{aligned} \]That is 5.3 \(\mu\)m/s, which rounds to 5. Using \(h\) instead of \(\hbar\) gives about 33, and leaving out the factor of 2 gives about 10.6, which is the trap behind the option 10.

Q: Out of the following particles, all having the same kinetic energy, which has the longest wavelength ?

The longest wavelength belongs to an electron.The de Broglie wavelength of a particle with kinetic energy K is\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} \]At the same kinetic energy, \(\lambda \propto 1/\sqrt{m}\), so the lightest particle has the longest wavelength. The electron is about 1836 times lighter than a proton, so its wavelength is about 43 times longer. The proton and neutron have nearly equal masses and wavelengths, and the alpha particle, about four times heavier still, has the shortest.

A student sets up Young’s double slit experiment with electrons of momentum \(p\) incident normally on the slits of width \(w\) separated by distance \(d\). In order to observe interference fringes on a screen at a distance \(D\) from the slits, which of the following conditions should be satisfied?

A.\(\frac{\hbar}{p} \gt \frac{Dw}{d}\)
B.\(\frac{\hbar}{p} \gt \frac{dw}{D}\)
C.\(\frac{\hbar}{p} \gt \frac{d^{2}}{D}\)
D.\(\frac{\hbar}{p} \gt \frac{d^{2}}{\sqrt{Dw}}\)

Exam Year(s)

GATE Physics - 2022,

Answer: B

The condition is \(\frac{\hbar}{p} \gt \frac{dw}{D}\).

The electrons have de Broglie wavelength \(\lambda = h/p\), of order \(\hbar/p\). The interference fringes on the screen are spaced by about \(\lambda D/d\). Each slit has a finite width \(w\), and electrons passing through different parts of one slit produce fringe patterns shifted relative to each other; the fringes stay visible only if their spacing is larger than this spread, that is roughly

\[ \begin{aligned} \frac{\lambda D}{d} &\gt w \\ \Rightarrow \frac{\hbar}{p} &\gt \frac{dw}{D} \end{aligned} \]

up to a numerical factor. The other options place \(D\) and \(d\) wrongly: \(Dw/d\) would demand a longer wavelength as the screen moves away, which is the reverse of the real dependence, and the forms with \(d^{2}\) do not involve the slit width at all.

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Young’s double slit experiment is performed using a beam of C₆₀ (fullerene) molecules, each molecule being made up of 60 carbon atoms. When the slit separation is 50 nm, fringes are formed on a screen kept at a distance of 1 m from the slits. Now, the experiment is repeated with C₇₀ molecules with a slit separation of 92.5 nm. The kinetic energies of both the beams are the same. The position of the 4^th bright fringe for C₆₀ will correspond to the \(n^{th}\) bright fringe for C₇₀. What is the value of \(n\) (rounded off to the nearest integer)?

Exam Year(s)

GATE Physics - 2023,

Answer: D

The value is \(n = 8\).

The \(n\)th bright fringe lies at \(y = n\lambda D/d\). With equal kinetic energies, \(\lambda = h/\sqrt{2mK}\), so \(\lambda \propto 1/\sqrt{m}\) and \(\lambda_{60}/\lambda_{70} = \sqrt{70/60}\). Equating the positions,

\[ \begin{aligned} \frac{4\lambda_{60}}{50} &= \frac{n\lambda_{70}}{92.5} \\ n &= 4 \times \frac{92.5}{50} \times \sqrt{\frac{70}{60}} \\ &= 7.4 \times 1.080 \approx 7.99 \end{aligned} \]

which rounds to 8. Ignoring the mass dependence of the wavelength gives 7.4, rounding to 7; using the mass ratio without the square root gives about 8.6. 5 and 6 do not follow from the fringe condition.

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A quantum particle of mass \(10^{-20}\) kg is confined within a length of 1 \(nm\) in one-dimension. The minimum uncertainty in the measurement of velocity of the particle, in units of \(\mu m/s\), rounded off to the nearest integer is:

[Assume the minimum uncertainty product \(\Delta x \Delta p_{x} \approx \frac{\hbar}{2}\) , use Planck’s constant h = \(6.64 \times 10^{-34} J\ s\)]

A.2
B.5
C.10
D.1

Exam Year(s)

IIT JAM - 2026,

Answer: B

The minimum velocity uncertainty is about 5 \(\mu m/s\).

With \(\Delta x = 1\) nm and \(\Delta p_x = m\,\Delta v\), the minimum uncertainty product gives

\[ \begin{aligned} \Delta v &= \frac{\hbar}{2m\,\Delta x}, \quad \hbar = \frac{6.64 \times 10^{-34}}{2\pi} \\ &= \frac{1.057 \times 10^{-34}}{2 \times 10^{-20} \times 10^{-9}} \\ &= 5.3 \times 10^{-6}\ \text{m/s} \end{aligned} \]

That is 5.3 \(\mu\)m/s, which rounds to 5. Using \(h\) instead of \(\hbar\) gives about 33, and leaving out the factor of 2 gives about 10.6, which is the trap behind the option 10.

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Out of the following particles, all having the same kinetic energy, which has the longest wavelength ?

A.An \(\alpha\)-particle
B.An electron
C.A proton
D.A neutron

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The longest wavelength belongs to an electron.

The de Broglie wavelength of a particle with kinetic energy K is

\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} \]

At the same kinetic energy, \(\lambda \propto 1/\sqrt{m}\), so the lightest particle has the longest wavelength. The electron is about 1836 times lighter than a proton, so its wavelength is about 43 times longer. The proton and neutron have nearly equal masses and wavelengths, and the alpha particle, about four times heavier still, has the shortest.

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