Question

If the radius of the first Bohr orbit is $r$, then the de Broglie wavelength of the electron in the $4^{\text{th}}$ orbit is

• A. $\omega\cos\phi$
• B. $a\omega$
• C. $\omega\cos2\phi$
• D. $-a\omega\cos2\phi$

Hint:

Physics Tip : In SHM, $v=\dfrac{dx}{dt}$ and maximum velocity is $a\omega$.

Answer 1

The Correct Option is B

Step 1: Given displacement equation. $$ x=a\sin(\omega t-\phi) $$ For S.H.M., velocity is derivative of displacement.

Step 2: Differentiate with respect to time. $$ v=\frac{dx}{dt} $$ $$ v=a\omega\cos(\omega t-\phi) $$

Step 3: Substitute given time. Given: $$ t=\frac{\phi}{\omega} $$ So, $$ v=a\omega\cos\left(\omega\cdot\frac{\phi}{\omega}-\phi\right) $$

Step 4: Simplify angle. $$ v=a\omega\cos(\phi-\phi) $$ $$ v=a\omega\cos0 $$ Since $\cos0=1$, $$ v=a\omega $$

Step 5: Conclusion. $$ \therefore \text{Correct option is (B).} $$