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JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter

'm' દળ ધરાવતો એક ઇલેક્ટ્રોન, જેનો પ્રારંભિક વેગ \(\overrightarrow{\mathrm{v}}=\mathrm{v}_0 \hat{i}\left(\mathrm{v}_0\gt0\right)\) છે, તે વિદ્યુતક્ષેત્ર \(\overrightarrow{\mathrm{E}}=-\mathrm{E}_{\mathrm{o}} \hat{\mathrm{k}}\) માં પ્રવેશે છે. જો પ્રારંભિક દ-બ્રોગ્લી તરંગલંબાઈ \(\lambda_0\) હોય, તો સમય t પછી તેનું મૂલ્ય શું હશે?

  1. A \(\frac{\lambda_0}{\sqrt{1+\frac{\mathrm{e}^2 \mathrm{E}_0{ }^2 \mathrm{t}^2}{\mathrm{~m}^2 \mathrm{v}_0{ }^2}}}\)
  2. B \(\lambda_{\mathrm{o}} \sqrt{1+\frac{\mathrm{e}^2 \mathrm{E}_0^2 \mathrm{t}^2}{\mathrm{~m}^2 \mathrm{v}_{\mathrm{o}}^2}}\)
  3. C \(\frac{\lambda_0}{\sqrt{1-\frac{\mathrm{e}^2 \mathrm{E}_{\mathrm{o}}{ }^2 \mathrm{t}^2}{\mathrm{~m}^2 \mathrm{v}_{\mathrm{o}}^2}}}\)
  4. D \(\lambda_0\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{\lambda_0}{\sqrt{1+\frac{\mathrm{e}^2 \mathrm{E}_0{ }^2 \mathrm{t}^2}{\mathrm{~m}^2 \mathrm{v}_0{ }^2}}}\)

Step-by-step Solution

Detailed explanation

\begin{aligned} & \overrightarrow{\mathrm{v}}=\mathrm{v}_0 \hat{\mathrm{i}}-\frac{\mathrm{E}_0 \mathrm{e}}{\mathrm{m}} t \hat{\mathrm{k}} \\ & |\overrightarrow{\mathrm{v}}|=\sqrt{\mathrm{v}_0^2+\frac{\mathrm{E}_0^2 \mathrm{e}^2 \mathrm{t}^2}{\mathrm{~m}^2}} \\ &…

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