Which of the following has the largest de Bro | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The de Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass,and $v$ is the velocity.

Vedclass · Accepted Answer

$A$ ball of mass $0.03 \ kg$ moving with speed $1 \ m/s$

Vedclass · Answer

(D) The de Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass,and $v$ is the velocity.Since $\lambda \propto \frac{1}{mv}$,the particle with the smallest momentum $(mv)$ will have the largest de Broglie wavelength.Calculating momentum $(p = mv)$ for each case:$A$: $p = 0.02 \ kg 	imes 1000 \ m/s = 20 \ kg \cdot m/s$$B$: $p = 0.06 \ kg 	imes 10 \ m/s = 0.6 \ kg \cdot m/s$$C$: $p = 0.01 \ kg 	imes 100 \ m/s = 1 \ kg \cdot m/s$$D$: $p = 0.03 \ kg 	imes 1 \ m/s = 0.03 \ kg \cdot m/s$Comparing the values,the momentum in option $D$ is the smallest $(0.03 \ kg \cdot m/s)$.Therefore,the ball in option $D$ has the largest de Broglie wavelength.

Which of the following has the largest de Broglie wavelength?

Similar Questions

An electron of mass $m$ with an initial velocity $\vec{V} = V_0 \hat{i} \,(V_0 > 0)$ enters an electric field $\vec{E} = -E_0 \hat{i} \,(E_0 = \text{constant} > 0)$ at $t = 0$. If $\lambda_0$ is its de-Broglie wavelength initially,then its de-Broglie wavelength at time $t$ is:

The de Broglie wavelength of an electron accelerated to a potential of $ 400 \,V $ is approximately (in $\,nm$)

The de Broglie wavelength of a proton and $\alpha$-particle are equal. The ratio of their velocities is ...... .

The additional energy that should be given to an electron to reduce its de-Broglie wavelength from $1 \ nm$ to $0.5 \ nm$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module