If numerical value of mass and velocity are equal then de-Broglie wavelength in terms of $\mathrm{KE}$ is

Question

Aicharya · Accepted Answer

Given that $\mathrm{m}=\mathrm{v}$ $\begin{aligned} & \lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\mathrm{~m}^2}---(1) \ & \mathrm{K} \cdot \mathrm{E}=\frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \mathrm{~m}^3---(2) \end{aligned}$ multiplying and deviding equation (1) by "m" $\lambda=\frac{\mathrm{hm}}{\mathrm{~m}^3}---(3)$ substituting $\mathrm{m}^3$ from equation (2) in equation (3) $\lambda=\frac{\mathrm{hm}}{2 \mathrm{~K} \cdot \mathrm{E}}$

If numerical value of mass and velocity are equal then de-Broglie wavelength in terms of $\mathrm{KE}$ is

Solution