JEE Main 2026 — Dual Nature of Matter Question with Solution

The de Broglie wavelength of an electron in free space is given by ${\lambda }_0=\frac{h}{mv}$.

When the electron enters the medium, its velocity is reduced by $20\%$. The new velocity is $v^{\prime }=v-0.20v=0.80v$.

The de Broglie wavelength in the medium is ${\lambda }^{\prime }=\frac{h}{m{v}^{\prime }}$.

Substituting $v^{\prime }=0.80v$, we get ${\lambda }^{\prime }=\frac{h}{m(0.80v)}=\frac{1}{0.80}\left(\frac{h}{mv}\right)$.

Since \dfrac{1}{0.80} = \dfrac{100}{80} = 1.25$,wehave$\lambda' = 1.25\lambda_0$.<br><br>Comparingthiswith$\lambda' = \alpha\lambda_0$,weget$\alpha = 1.25$.<br><br>Answer:$1.25$