11.5 The energy flux of sunlight reaching the surface of the earth is 1.388×103 W/m2. How many photons (nearly) per square meter are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.
11.5 The energy flux of sunlight reaching the surface of the earth is 1.388×103 W/m2. How many photons (nearly) per square meter are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.
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1 Answer
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11.5 The energy flux of sunlight, = 1.388 W/
Hence power of the sunlight per square meter, P = 1.388
Speed of light, c = 3 m/s
Planck’s constant, h = 6.626 Js
Average wavelength of photon, = 550 nm = 550 m
If n is the number of photon per square meter, incident on earth per second, the equation of power can be written as
or =
We know =
Hence, = = = 3.84 photons /s
Therefore, every second 3.84 photons are incident per square meter on earth.
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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
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
This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:
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