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Electromagnetic Waves question

2021 · 25 Feb · Shift 2 · Q73
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  5. /2021 · 25 Feb · Shift 2 · Q73

Electromagnetic Waves question

2021 · 25 Feb · Shift 2 · Q73

JEE MainPhysicsElectromagnetic WavesNumerical+4 / −1
The wavelength of an X-ray beam is 10 Ao\mathop A\limits^oAo​. The mass of a fictitious particle having the same energy as that of the X-ray photons is x3h{x \over 3}h3x​h kg. The value of x is ‾\underline{\hspace{2cm}}​. (h = Planck's constant)
Numerical answer
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Correct answer: 10

  1. Energy of one X-ray photon

Given wavelength: λ=10 A˚=10×10−10 m=10−9 m\lambda = 10\,\text{\AA} = 10 \times 10^{-10}\,\text{m} = 10^{-9}\,\text{m}λ=10A˚=10×10−10m=10−9m

Energy of a photon is: E=hcλE = \frac{hc}{\lambda}E=λhc​

So, E=hc10−9=109hcE = \frac{hc}{10^{-9}} = 10^9 hcE=10−9hc​=109hc

  1. Energy equivalent mass of fictitious particle

If a particle has the same energy, then using Einstein's relation: E=mc2E = mc^2E=mc2

Hence, m=Ec2=hc/λc2=hλcm = \frac{E}{c^2} = \frac{hc/\lambda}{c^2} = \frac{h}{\lambda c}m=c2E​=c2hc/λ​=λch​

Substitute λ=10−9 m\lambda = 10^{-9}\,\text{m}λ=10−9m and c=3×108 m/sc = 3 \times 10^8\,\text{m/s}c=3×108m/s: m=h(10−9)(3×108)m = \frac{h}{(10^{-9})(3\times 10^8)}m=(10−9)(3×108)h​

m=h3×10−1=103hm = \frac{h}{3 \times 10^{-1}} = \frac{10}{3}hm=3×10−1h​=310​h

  1. Compare with given form

Given mass is: m=x3hm = \frac{x}{3}hm=3x​h

So, x3h=103h\frac{x}{3}h = \frac{10}{3}h3x​h=310​h

Therefore, x=10x = 10x=10

Final Answer

10\boxed{10}10​

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