Class 11th Structure of Atom Chemistry

**A 662 watt bulb emits monochromatic light of wavelength 300 nm. The number of photons emitted per second by the bulb is (take $h = 6.62 * 10^{- 34} J s$ )**

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>21</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>s</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </mrow> </msup> </math> </span></p>

Option 2 - <p><span class="mathml" contenteditable="false"> <math> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>20</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>s</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </mrow> </msup> </math> </span></p>

Option 3 - <p><span class="mathml" contenteditable="false"> <math> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>18</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>s</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </mrow> </msup> </math> </span></p>

Option 4 - <p><span class="mathml" contenteditable="false"> <math> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>19</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>s</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </mrow> </msup> </math> </span></p>

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Vishal Baghel | Contributor-Level 10

4 months ago

Correct Option - 1

Detailed Solution:

Power of bulb = 662 watt = 662 Js^-1

Energy of one photon $= \frac{6.62 * 10^{- 34} * 3 * 10^{8}}{300 * 10^{- 9}}$ $\frac{^{}^{}}{^{}}$

$= 6.62 * 10^{- 19} J$

Number of photons emitted $= \frac{662}{6.62 * 10^{- 19}}$ $\frac{}{^{}}$

$= 10^{21} s^{- 1}$

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