There are two sources of light, each emitting with a power of 100 W. One emits X-rays of wavelength 1 nm and the other visible light at 500 nm. Find the ratio of number of photons of X-rays to the photons of visible light of the given wavelength.

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7 months ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- p=100W

λ 1= 1nm

λ 2= 500nm

Also n1and n2 represent number of photon of x-rays and visible light emitted from the two sources per sec

E/t=P=n1hc/ λ 1=n2hc/ λ 2

So n1/ λ 1 = n2/ λ 2

So n1/n2= 1/500

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This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

f(x)=2x23x2x2=2x24x+x2x2=2x(x2)+1(x2)x2=(2x+1)(x2)x2=2x+1limx2f(x)=2x+1=limh02(2h)+1=4+1=5limx2+f(x)=2x+1=limh02(2+h)+1=4+1=5limx2f(x)=5Aslimx2f(x)=limx2+f(x)=limx2f(x)=5Hence,f(x)iscontinuousatx=2.

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Physics Ncert Solutions Class 12th 2023

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