Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
An electron spends about 10^-8s in an excited state before it drops to a lower state by emiting a photon. How many revolution does an electron in an n=2 bohr orbit make in 1.00*10^-8s? (ans_8.23*10^6 revolution)
velocity in any orbit = (2.18*106)/n m/s V2 = (2.18*106)/2 m/s radius of any orbit = (52.9)*10-12n2 m r2 = (52.9*10-12)4 m time in which electron completes one revolution is T = 2pir2/V2 T =12*10-16 let it completes n crevolution so total time taken = n(12*10-16) n(12*10-16) = 10-8 n = 8.23 * 106 revolutions
velocity in any orbit = (2.18*106)/n m/s
V2 = (2.18*106)/2 m/s
radius of any orbit = (52.9)*10-12n2 m
r2 = (52.9*10-12)4 m
time in which electron completes one revolution is
T = 2pir2/V2
T =12*10-16
let it completes n crevolution so total time taken = n(12*10-16)
n(12*10-16) = 10-8
n = 8.23 * 106 revolutions
Find solution of question on doubt solving app Instasolv
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -