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  • The de Broglie wavelength and kinetic energy of a particle is and 1 <img alt="\mathrm{eV}" src="https://ent

The de Broglie wavelength and kinetic energy of a particle is 2000 \AA and 1 \mathrm{eV} respectively. If its kinetic energy becomes 1\mathrm{~eV}, then its de Broglie wavelength becomes
 

Option: 1

1 \AA


Option: 2

5 \AA


Option: 3

2 \AA


Option: 4

10 \AA


Answers (1)

best_answer

As \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}

Since mass of the particle remains constant

\begin{aligned} & \therefore \lambda \propto \frac{1}{\sqrt{\mathrm{K}}} \\ & \frac{\lambda^{\prime}}{\lambda}=\sqrt{\frac{\mathrm{K}}{\mathrm{K}^{\prime}}}=\sqrt{\frac{1 \mathrm{eV}}{1 \times 10^6 \mathrm{eV}}}=\frac{1}{10^3} \\ & \text { or } \quad \lambda^{\prime}=\frac{\lambda}{10^3}=\frac{2000}{10^3} \AA=2 \AA \end{aligned}

Posted by

seema garhwal

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