Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
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 )
If the average life time of an excited state of H atom is of order 10 -8 sec, estimate how many orbits an e - makes when it is in the state n=2 and before it suffers a transition to n=1 state
speed of electron in any orbit = vn = vo/n (vo = 2.18*106m/s) so , v2 = vo/2 = 1.09*106m/s radius of orbit = rn = aon2 (ao = 52.9 * 10-12m) so , r2 = 2.12*10-10m v = X/T x = @r so ( r is radius of second orbit) v = @r/T ( @ is angular displacement) @ = vT/r @ = 2pi(n) (here n is number of revolutions) 2pi(n) = vT/r .......................1 substituting values of v,r,T we get n = 8.18*106 revolutions approve if u like this ans
speed of electron in any orbit = vn = vo/n (vo = 2.18*106m/s)
so , v2 = vo/2 = 1.09*106m/s
radius of orbit = rn = aon2 (ao = 52.9 * 10-12m)
so , r2 = 2.12*10-10m
v = X/T
x = @r so ( r is radius of second orbit)
v = @r/T ( @ is angular displacement)
@ = vT/r
@ = 2pi(n) (here n is number of revolutions)
2pi(n) = vT/r .......................1
substituting values of v,r,T we get
n = 8.18*106 revolutions
approve if u like this ans
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -