Solution:We know that
E0=−13.6n2eVatom for atom
Let electron jump from n2 to n1
then
En2−En1=13.6n22−13.6n22
⟹2.55=13.61n21−1n22
By hit and trial we get n2=4 and n1=2 [angular momentum m v r = nh2π]
Change in angular momentum = m1h2π−m2h2π=h2π(2−4)=h2π×(−2)=hπ
The momentum of emitted photon can be found by de Broge relationship
λhp ⟹P=hλ=hvc=Ec
∴2.55×1.6×10−193×108=0.8m/sec
Hence B is the correct option