Question

If a 60 watt lamp emits monochromatic
radiation of wavelength 6625 Å, then the
number of photons emitted per second by
the lamp is (h = 6.625 x 10-34 Js)​

Answer 1

Answer:Is 20 number of photons passed per second

Explanation:By E=Nhv

E=Nhv/lembda

E×lembda/hv

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Answer 2

Answer: 5×10⁻²⁰

Given:

$\sf Power\:of\:the\:lamp\:(P)=60\:W$

$\sf Wavelength\:of\:the\:lamp(\lambda)=6625\AA=6625\times10^{-10}=6.625\times10^{-7}m$

$\sf Planck's\:constant(h)=6.625\times 10^{-34}Js$

To find:

$\sf Number\:of\:photons\:emitted\:per\:second\:by\:the\:lamp(n).$

Formula used:

$\sf\bullet \boxed{\sf Energy\:of\:photon(E)=Planck's\:constant(h)\times Frequency(\nu)}$

$\bullet\boxed{\sf Power(P)=Number\:of\:photons\:emitted\:per\:second(n)\times Energy\:of\:photon(E)}$

Explanation:

$\sf We\:know\:that,$

$\sf Speed\:of\:light\:in\:vaccum(c)=\lambda \nu$

$\Rightarrow \sf\nu=\dfrac{c}{\lambda}\quad\quad\quad (where\:c=3\times 10^{8})$

$\sf Substituting\:the\:value\:of\:\nu\:in\:the\:formula,we\:get:$

$\boxed{\sf E=\dfrac{hc}{\lambda}}$

$\sf Now, substituting\:the\:values\:in\:the\:above\:formula,we\:get:$

$\Rightarrow \sf E=\dfrac{ (6.625\times 10^{-34})\times (3\times 10^{8})}{6.625\times 10^{-7}}$

$\Rightarrow \sf E= 10^{-27}\times 3\times 10^{8}$

$\Rightarrow \sf E= 3\times 10^{-19}$

$\sf Again\:substituting\:the\:values\:in\:the\:formula,we\:get:$

$\Rightarrow\sf 60=n\times( 3\times 10^{-19})$

$\Rightarrow\sf n=\dfrac{ 3\times 10^{-19}}{60}=\dfrac{ 1\times 10^{-19}}{20}=0.05\times 10^{-19}$

$\Rightarrow\boxed{\sf n=5\times 10^{-20}\:photons\:per\:second}$

Therefore,the  number of photons emitted per second by

the lamp is 5×10⁻²⁰.