CUET · MATHS · PYQ PAPER 2024
A proton accelerated through potential difference V has de Broglie wavelength \(\lambda\). On doubling the potential, the de Broglie wavelength of the proton :
- A remains unchanged
- B becomes double
- C becomes four times
- D decreases
Answer & Solution
Correct Answer
(D) decreases
Step-by-step Solution
Detailed explanation
\(\lambda = \frac{h}{\sqrt{2mqV}}\) \(\lambda' = \frac{h}{\sqrt{2mq(2V)}} = \frac{1}{\sqrt{2}} \frac{h}{\sqrt{2mqV}}\) \(\lambda' = \frac{\lambda}{\sqrt{2}}\) decreases
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