Question

The de-Broglie wavelength of an electron in first orbit of Bohr's hydrogen is equal to

• A. radius of the orbit
• B. perimeter of the orbit
• C. diameter of the orbit
• D. half of the perimeter of the orbit

Answer 1

The correct option is B perimeter of the orbit

According to the expression given by de-Broglie for the wavelength of a particle of momentum $mv$
$\lambda =\frac{h}{mv}$ -------(1)
According to the Bohr's postulate, electrons move only in those orbits where angular momentum is quantised.
$mvr=\frac{nh}{2\pi }$ ---------(2)
Putting the expression for $mv$ from equation 1 to equation 2,
we get $n\lambda =2\pi r$
For first orbit $n=1$
Hence $\lambda =2\pi r$
which is perimeter of the orbit.