An electron accelerated through a potential difference $V_{1}$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_{2}$, its de-Broglie wavelength increases by $50 \%$. The value of $\left(\frac{V_{1}}{V_{2}}\right)$ is equal to
Solution
<p>$P = \sqrt {2\,eVm}$</p>
<p>$\lambda = \left( {{h \over {{P_1}}}} \right)$ ..... (i)</p>
<p>${{3\lambda } \over 2} = {h \over {{P_2}}}$ ..... (ii)</p>
<p>Dividing (i) by (ii)</p>
<p>$$ \Rightarrow {2 \over 3} = \left( {{{{P_2}} \over {{P_1}}}} \right) = \sqrt {{{{v_2}} \over {{v_1}}}} $$</p>
<p>$\Rightarrow {4 \over 9} = \left( {{{{v_2}} \over {{v_1}}}} \right)$</p>
<p>${{{v_1}} \over {{v_2}}} = \left( {{9 \over 4}} \right)$</p>
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: de Broglie Hypothesis
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