When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is
When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes
If
the value of x will be __________.
When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes If the value of x will be __________.
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1 Answer
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hu = hu0 + K.E
Cases u = 2u0
h2u0 = hu0 + K.E1
K.E1 = hu0
- (1)
Now, cases 2
h 5u0 = hu0 + k.E2
k.E2 = 4hu0
v2 =
v2 = 2v1
Similar Questions for you
Based on theory
z² × (13.6) (1 - ¼) = 3 × (13.6)
z = 2 . (i)
h/√2mk? = (1/2.3) × h/√2mk?
=> k? = (2.3)²k? = 5.25k? (ii)
Now, k? = E? - Φ
k? = E? - Φ = z²E? - Φ
∴ k? /k? = (10.2 - Φ)/ (4 × 10.2 - Φ) = 1/5.25
=> Φ = 3eV
- (i)
- (ii)
from (i) & (ii)
ev
This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:
This is a Short Answer Type Questions as classified in NCERT Exemplar
Explanation- p=100W
1= 1nm
2= 500nm
Also n1and n2 represent number of photon of x-rays and visible light emitted from the two sources per sec
E/t=P=n1hc/ 1=n2hc/ 2
So n1/ 1 = n2/ 2
So n1/n2= 1/500
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