The de-Broglie wavelength of an electron in the first Bohr orbit is

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(d) : Angular momentum $=\frac{n h}{2 \pi}$ ⇒ moment of momentum $=p 	imes r_n$ $$ \Rightarrow p 	imes r_n=\frac{n h}{2 \pi} \Rightarrow \frac{h}{\lambda} r_n=\frac{n h}{2 \pi} \Rightarrow \lambda=\frac{2 \pi r_n}{n} $$ For $1^{	ext {st }}$ orbit, $n=1, \lambda=2 \pi r_1$ $\Rightarrow \lambda=$ circumference of $1^{	ext {st }}$ orbit.