The number of photons emitted by a monochromatic (single frequency) infrared range finder of power 1 mW and wavelength of 1000 nm, in 0.1 second is x $\times$ 1013. The value of x is _____________. (Nearest integer)
(h = 6.63 $\times$ 10$-$34 Js, c = 3.00 $\times$ 108 ms$-$1)
Answer (integer)
50
Solution
Energy emitted in 0.1 sec.<br><br>= 0.1 sec. $\times$ ${10^{ - 3}}{J \over s}$<br><br>= 10<sup>$-$4</sup> J<br><br>If 'n' photons of $\lambda$ = 1000 nm are emitted, then 10<sup>$-$4</sup> = n $\times$ ${{hc} \over \lambda }$<br><br>$$ \Rightarrow {10^{ - 4}} = {{n \times 6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {1000 \times {{10}^{ - 9}}}}$$<br><br>$\Rightarrow$ n = 5.02 $\times$ 10<sup>14</sup> = 50.2 $\times$ 10<sup>13</sup><br><br>$\Rightarrow$ 50 (nearest integer)
About this question
Subject: Chemistry · Chapter: Atomic Structure · Topic: Bohr's Model
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