Question 122: Dual Nature of Matter and Radiation

Question

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

Accepted Answer

$$P = \sqrt {2\,eVm} $$
$$\lambda  = \left( {{h \over {{P_1}}}} \right)$$ ..... (i)
$${{3\lambda } \over 2} = {h \over {{P_2}}}$$ ..... (ii)
Dividing (i) by (ii)
$$ \Rightarrow {2 \over 3} = \left( {{{{P_2}} \over {{P_1}}}} \right) = \sqrt {{{{v_2}} \over {{v_1}}}} $$
$$ \Rightarrow {4 \over 9} = \left( {{{{v_2}} \over {{v_1}}}} \right)$$
$${{{v_1}} \over {{v_2}}} = \left( {{9 \over 4}} \right)$$

Answer

$\frac{3}{2}$

Answer

4

Answer

3

Answer

$\frac{9}{4}$

Dual Nature of Matter and Radiation

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

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