11.Dual Nature of Radiation and matter
medium

એક ઉદ્‍ગમ $S_1$, પ્રતિ સેકન્ડે $5000\;\mathring A$ તરંગલંબાઈના $10^{15}$ ફોટોન ઉત્સર્જે છે. બીજો ઉદ્‍ગમ $S_2$ પ્રતિ સેકન્ડે $5100\;\mathring A$ તરંગલંબાઈના $1.02 \times 10^{15}$ ફોટોન ઉત્સર્જે છે. ($S_2$ ઉદ્‍ગમનો પાવર)/($S_1$ ઉદ્‍ગમનો પાવર કોને બરાબર થાય?

A

$1$

B

$1.02$

C

$1.04$

D

$0.98$

(AIPMT-2010)

Solution

For a source $S_{1}$

Wavelength, $\lambda_{1}=5000 \,\,{\mathop {\text{A}}\limits^o }$

Number of photons emitted per second, $N_{1}=10^{15}$

Energy of each photon, $E_{1}=\frac{h c}{\lambda_{1}}$ 

Power of source $S_{1}, P_{1}=E_{1} N_{1}=\frac{N_{1} h c}{\lambda_{1}}$

For a source $S_{2}$ 

Wavelength, $\lambda_{2}=5100\,\,{\mathop {\text{A}}\limits^o }$

Number of photons emitted per second,

$N_{2}=1.02 \times 10^{15}$

Energy of each photon, $E_{2}=\frac{h c}{\lambda_{2}}$

Power of source $S_{2},\,\,P_{2}=N_{2} E_{2}=\frac{N_{2} h c}{\lambda_{2}}$

$\therefore \,\,\frac{{{\text{ Power of}}\,{\text{ }}{S_2}}}{{{\text{ Power of }}\,{S_1}}} = \frac{{{P_2}}}{{{P_1}}}$ $ = \frac{{\frac{{{N_2}hc}}{{{\lambda _2}}}}}{{\frac{{{N_1}hc}}{{{\lambda _1}}}}} = \frac{{{N_2}{\lambda _1}}}{{{N_1}{\lambda _2}}}$

$ = \frac{{(1.02 \times {{10}^{15}}{\text{ photons/s) }} \times (5000{\mkern 1mu} {\mkern 1mu} \mathop {\text{A}}\limits^o )}}{{({{10}^{15}}{\text{ photons/s) }} \times (5100{\mkern 1mu} {\mkern 1mu} \mathop {\text{A}}\limits^o )}}$ $ = \frac{{51}}{{51}} = 1$

Standard 12
Physics

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