Question

How De Broglie wavelength is related to velocity for particles having same kinetic energy

Answer 1

Dear Student

$\lambda =\frac{h}{mv}$; h is Planck's constant, m is the mass, v is the velocity

$$\begin{gathered} \frac{{\lambda }_1}{{\lambda }_2}=\frac{{\displaystyle \frac{h}{m_1{v}_1}}}{{\displaystyle \frac{h}{m_2{v}_2}}}=\frac{m_2{v}_2}{m_1{v}_1} \\ Dividebothnumeratoranddenominatorby2andmultiplyby{v}_{2}and{v}_1 \\ \frac{{\lambda }_1}{{\lambda }_2}=\frac{{\displaystyle \frac{m_2{v}_2\times v_2\times v_1}{2}}}{{\displaystyle \frac{m_1{v}_1\times v_1\times v_2}{2}}} \end{gathered}$$

$$\begin{gathered} Kineticenergy,K=\frac{1}{2}m{v}^2 \\ \frac{{\lambda }_1}{{\lambda }_2}=\frac{{\displaystyle K_2\times v_1}}{{\displaystyle K_1\times v_2}} \\ {K}_2=K_1 \\ Hence,\frac{{\lambda }_1}{{\lambda }_2}=\frac{{\displaystyle v_1}}{{\displaystyle v_2}} \\ \lambda \alpha v \end{gathered}$$

De Broglie wavelength is directly proportional to velocity for particles having same kinetic energy​​​​​​

Regards