Question

The number of photons emitted in 10 hour by a 40 W sodium lamp (N = 5890 Å) is 0 6.41 X 1024 0 4.26 x 1021 O 4.26 x 1024 6.4 x 1020​

Answer 1

Answer:

Correct option is

A

6.5×10

24

E=P×t

=60×10×60×60

=216×10

4

J

λ=6000

A

˚

=6000×10

−10

m

E=

λ

nhc

n=

hc

Eλ

=

6.626×10

−34

×3×10

8

216×10

4

×6000×10

−10

=6.5×10

24

photons.

Answer 2

The number of photon emitted is 4.26 × 10²⁴

Therefore, option (3) is correct.

Explanation:

Given:

Power of Sodium lamp = 40 W

Duration = 10 hours

Wavelength of light λ = 5890 Å

To find out:

The number of photons emitted

Solution:

Given

$P=40$ W

Energy dissipated in 10 hours

$E=P\times t$

$\implies E= 40\times 10\times 60\times 60$

$\implies E=1.44\times 10^6$ Jule

Energy of one photon

$E_p=h\nu$

$\implies E_p=h\frac{c}{\lambda}$

$\implies E_p=\frac{6.6\times 10^{-34}\times 3\times 10^8}{5890\times 10^{-10}}$

$\implies E_p=3.36\times 10^{-18}$

Thus, the number of photons emitted

$n=\frac{E}{E_p}$

$\implies n=\frac{1.44\times 10^6}{3.36\times 10^{-18}}$

$\implies n=4.26\times 10^{24}$

Hope this answer is helpful.

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