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Chemistry
If kinetic energy of a proton is increased, nine times the wavelength of the de Broglie wave associated with it would become
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Q. If kinetic energy of a proton is increased, nine times the wavelength of the de Broglie wave associated with it would become
Structure of Atom
A
$3$ times
B
$9$ times
C
$\frac{1}{3}$ times
D
$\frac{1}{9}$ times.
Solution:
$\lambda=\frac{h}{\sqrt{2 m E}}$
$\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{E_{2}}{E_{1}}}$
$=\sqrt{\frac{9 E_{1}}{E_{1}}}=3$
$\lambda_{2}=\frac{1}{3} \lambda_{1}$