A proton,an electron,and an alpha particle ha | vedclass

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

Question

(A) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant,$m$ is the mass,and $K$ is the

Vedclass · Accepted Answer

$\lambda_{e} > \lambda_{p} > \lambda_{\alpha}$

Vedclass · Answer

(A) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant,$m$ is the mass,and $K$ is the kinetic energy.Since the kinetic energies $(K)$ are the same for all particles,$\lambda \propto \frac{1}{\sqrt{m}}$.We know the masses of the particles are: $m_{e} Since $\lambda$ is inversely proportional to the square root of mass,the order of wavelengths will be $\lambda_{e} > \lambda_{p} > \lambda_{\alpha}$.

$A$ proton,an electron,and an alpha particle have the same kinetic energies. Their de-Broglie wavelengths will be compared as:

Similar Questions

The kinetic energy of a free electron increases to $3$ times the previous kinetic energy $(K.E.)$. The ratio of the new de-Broglie wavelength to the previous de-Broglie wavelength is:

State the de-Broglie hypothesis.

The de-Broglie hypothesis treated electrons as

An electron is accelerated through a potential difference of $10,000 \ V$. Its de-Broglie wavelength is nearly: $(m_{e} = 9 \times 10^{-31} \ kg)$

In an optical microscope, which electromagnetic waves are used? Which type of waves are used in an electron microscope?

Vedclass Test Series

Exam Paper Generator

Online Exam Module