Electrons are accelerated to a speed of $4.0 \times 10 ^ { 4 } \mathrm {~m} / \mathrm { s }$ and are then aimed at a double-slit apparatus, where interference fringes are detected. If the electrons were replaced by neutrons, what speed must neutrons have to produce interference fringes with the same fringe spacing as that observed with the electrons? $\left( h = 6.626 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm { s } , m _ { \text {electron } } = 9.11 \times 10 ^ { - 31 } \mathrm {~kg} , m _ { \text {neutron } } = 1.675 \times 10 ^ { - 27 } \right.$ $\mathrm { kg }$ )
A) $2.2 \times 10 ^ { 1 } \mathrm {~m} / \mathrm { s }$
B) $9.3 \times 10 ^ { 2 } \mathrm {~m} / \mathrm { s }$
C) $7.3 \times 10 ^ { 7 } \mathrm {~m} / \mathrm { s }$
D) $1.7 \times 10 ^ { 6 } \mathrm {~m} / \mathrm { s }$

Question

Quizplus · Accepted Answer

The de Broglie wavelength is given by $\lambda = \frac{h}{p}$, where $h$ is Planck's constant and $p$ is the momentum of the particle. For the electrons, the momentum is $p = mv$, where $m$ is the mass of the electron and $v$ is its speed. For the neutrons, the momentum is $p = mv$, where $m$ is the mass of the neutron and $v$ is its speed. For the interference fringes to have the same spacing, the de Broglie wavelengths of the electrons and neutrons must be the same. Therefore, we have:$$\lambda_e = \lambda_n$$$$\frac{h}{p_e} = \frac{h}{p_n}$$$$p_e = p_n$$$$mv_e = mv_n$$$$v_n = v_e\frac{m_e}{m_n}$$$$v_n = (4.0 \times 10^4 \mathrm{~m/s})\frac{9.11 \times 10^{-31} \mathrm{~kg}}{1.675 \times 10^{-27} \mathrm{~kg}}$$$$v_n = 2.2 \times 10^1 \mathrm{~m/s}$$

Electrons Are Accelerated to a Speed Of 4.0×104 m/s4.0 \times 10 ^ { 4 } \mathrm {~m} / \mathrm { s }4.0×104 m/s

Electrons Are Accelerated to a Speed Of $4.0 \times 10 ^ { 4 } \mathrm {~m} / \mathrm { s }$