The angular momentum for the electron in Bohr’s orbit is L . If the electron is assumed to revolve in…

Bohr's formula for angular momentum of an electron in an orbit is given by 

$L=\frac{nh}{2\pi }$

For the first orbit, $n=1$ and the angular momentum is $L_1=\frac{h}{2\pi }=L$

For $2^{\mathrm{nd}}$ orbit

$L_2=\frac{2h}{2\pi }=2L$

Hence, the change in angular momentum is 

$\Delta L=2L-L=L$