The number of photons falling per second on a completely darkened plate to produce a force of 6.62 × 10-5 N is ‘n' If I the wavelength of the light falling is 5 × 10-7 m, then n = .... × 1022. (h = 6.62 × 10- 34 J-s)

Question

The number of photons falling per second on a completely darkened plate to produce a force of 6.62 × 10-5 N is ‘n' If I the wavelength of the light falling is 5 × 10-7 m, then n = .... × 1022. (h = 6.62 × 10- 34 J-s) (A) 1 (B) 5 (C) 0.2 (D) 3.3

Tardigrade Admin · Accepted Answer

P = (n/t) . (hc/λ) FC = (n/t) . (hC/λ) (n/t) = (F λ/h) (n/t) = (6.62 ×10-5 ×5 ×10-7/6.62 ×10-34) = 5×1022

Q. The number of photons falling per second on a completely darkened plate to produce a force of $6.62 \times 1 0^{- 5} N$ is $‘ n^{'}$ If I the wavelength of the light falling is $5 \times 1 0^{- 7} m,$ then $n = .... \times 1 0^{22}$ . (h = $6.62 \times 1 0^{- 34}$ J-s)

1

5

0.2

3.3

$P = \frac{n}{t} . \frac{h c}{λ}$
$FC = \frac{n}{t} . \frac{h C}{λ}$
$\frac{n}{t} = \frac{F λ}{h}$
$\frac{n}{t} = \frac{6.62 \times 1 0 ^{- 5} \times 5 \times 1 0 ^{- 7}}{6.62 \times 1 0 ^{- 34}}$
$= 5 \times 1 0^{22}$

Q. The number of photons falling per second on a completely darkened plate to produce a force of 6.62×10−5N is ‘n′ If I the wavelength of the light falling is 5×10−7m, then n=....×1022. (h = 6.62×10−34 J-s)

1

5

0.2

3.3

P=tn​.λhc​ FC=tn​.λhC​ tn​=hFλ​ tn​=6.62×10−346.62×10−5×5×10−7​ =5×1022

Q. The number of photons falling per second on a completely darkened plate to produce a force of $6.62 \times 1 0^{- 5} N$ is $‘ n^{'}$ If I the wavelength of the light falling is $5 \times 1 0^{- 7} m,$ then $n = .... \times 1 0^{22}$ . (h = $6.62 \times 1 0^{- 34}$ J-s)

$P = \frac{n}{t} . \frac{h c}{λ}$
$FC = \frac{n}{t} . \frac{h C}{λ}$
$\frac{n}{t} = \frac{F λ}{h}$
$\frac{n}{t} = \frac{6.62 \times 1 0 ^{- 5} \times 5 \times 1 0 ^{- 7}}{6.62 \times 1 0 ^{- 34}}$
$= 5 \times 1 0^{22}$