The angular momentum for the electron in Bohr’s orbit is L ....

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

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

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

For ${2}^{nd}$ orbit

${L}_{2} = \dfrac{2h}{2\pi} = 2L$

Hence, the change in angular momentum is

$\DeltaL = 2L - L = L$