11.23 (a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Å. What is the maximum energy of a photon in the radiation?

(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?

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8 months ago

11.23 Wavelength produced by X-ray, λ= 0.45 Å = 0.45 *10-10 m

Planck's constant, h = 6.626 *10-34 Js

Speed of light, c = 3 *108 m/s

The maximum energy of a photon is given as:

E = hcλ = 6.626*10-34*3*1080.45*10-10 = 4.417 *10-15 J = 4.417*10-151.6*1019 eV = 27.6 *103 eV = 27.6 keV

Therefore, the maximum energy of an X-ray photon is 27.6 keV

To get an X-ray of 27.6 keV, the incident

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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

Physics Ncert Solutions Class 12th 2023

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