# If the momentum of an electron is increased by Delta p, then the de Broglie wavelength associated with it changes by 0.50%. The initial momentum of the electron will be: A Delta p/200 B Delta p/199 C 199 Delta p D 400 Delta p

## Question

If the momentum of an electron is increased by $\Delta p$, then the de Broglie wavelength associated with it changes by 0.50%. The initial momentum of the electron will be:

A $\frac{\Delta p}{200}$ B $\frac{\Delta p}{199}$ C $199 \Delta p$ D $400 \Delta p$

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## Answer

The correct option is **C**: $199 \Delta \mathrm{p}$.

We start with the de Broglie wavelength formula: $$ \lambda = \frac{h}{p} $$ When the momentum is increased by $\Delta p$, the new wavelength becomes: $$ \lambda' = \frac{h}{p + \Delta p} $$

Given that the wavelength changes by **0.5%**, we can express the new wavelength as:
$$ \lambda' = \left(1 - \frac{0.5}{100}\right) \lambda = \frac{199}{200} \lambda $$

Substituting the new wavelength formula into the de Broglie relationship: $$ \frac{199 \lambda}{200} = \frac{h}{p + \Delta p} $$

Rewriting the original wavelength formula $\lambda = \frac{h}{p}$, we substitute $\lambda$: $$ \frac{199 h}{200 p} = \frac{h}{p + \Delta p} $$

By cross-multiplying to eliminate $h$: $$ \frac{199}{200 p} = \frac{1}{p + \Delta p} $$

Solving for $p + \Delta p$: $$ p + \Delta p = \frac{200}{199} p $$

Finally, isolating ( p ): $$ \Delta p = \frac{200}{199} p - p = \frac{1}{199} p $$

So, $$ p = 199 \Delta \mathrm{p} $$

Thus, the initial momentum of the electron is **$199 \Delta \mathrm{p}$**.

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