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JEE Main
Exam
JEE Main
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main · Physics · Electromagnetic Waves

X-rays of wavelength 0.140 ~nm are scattered from a block of...

X-rays of wavelength 0.140 nm0.140 \mathrm{~nm} are scattered from a block of carbon. What will be the wavelengths of XX-rays scattered at 90∘90^{{}^{{}^{\circ}}} ?

Options

  1. A0.140 nm0.140 \mathrm{~nm}
  2. B0.142 nm0.142 \mathrm{~nm}
  3. C0.144 nm0.144 \mathrm{~nm}
  4. D0.146 nm0.146 \mathrm{~nm}

Correct answer

B. 0.142 nm0.142 \mathrm{~nm}

Step-by-step solution

For ϕ=90∘,cos⁡ϕ=0\phi=90^{{}^{{}^{\circ}}}, \cos \phi=0
So,
<br/>λ′=λ+hmec<br/>=0.140×10−9+6.63×10−34(9.1×10−31)(3×108)<br/>=(0.140×10−9+2.4×10−12)m<br/>=0.142 nm<br/>\begin{aligned}<br/>& \lambda^{\prime}=\lambda+\frac{h}{m_e c} \\<br/>& =0.140 \times 10^{-9}+\frac{6.63 \times 10^{-34}}{\left(9.1 \times 10^{-31}\right)\left(3 \times 10^8\right)} \\<br/>& =\left(0.140 \times 10^{-9}+2.4 \times 10^{-12}\right) \mathrm{m} \\<br/>& =0.142 \mathrm{~nm}<br/>\end{aligned}
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