X-rays of wavelength 0.140 nm are scattered | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The Compton shift formula is given by $\Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c} (1 - \cos \phi)$.Given $\lambda = 0.140 \ nm = 0.140

Vedclass · Accepted Answer

$0.142$

Vedclass · Answer

(B) The Compton shift formula is given by $\Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c} (1 - \cos \phi)$.Given $\lambda = 0.140 \ nm = 0.140 	imes 10^{-9} \ m$ and $\phi = 90^{\circ}$.Since $\cos 90^{\circ} = 0$,the shift is $\Delta \lambda = \frac{h}{m_e c}$.The Compton wavelength $\frac{h}{m_e c} \approx 2.426 	imes 10^{-12} \ m = 0.002426 \ nm$.Therefore,$\lambda' = \lambda + \Delta \lambda = 0.140 \ nm + 0.002426 \ nm = 0.142426 \ nm$.Rounding to three decimal places,we get $\lambda' \approx 0.142 \ nm$.

$X$-rays of wavelength $0.140 \ nm$ are scattered from a block of carbon. What will be the wavelength of $X$-rays scattered at $90^{\circ}$ (in $nm$)?

Similar Questions

The product of the perpendicular distances from the origin to the pair of straight lines $12x^2+25xy+12y^2+10x+11y+2=0$ is

Haemozoin is a

The area (in sq units) of the triangle formed by the tangent at $(\sqrt{3}, 0)$ to the hyperbola $x^2-3y^2=3$ with the pair of asymptotes of the hyperbola is

The current through an ideal $PN$-junction diode shown in the following circuit diagram will be ..... $mA$.

Let $f(x) = x^3 + bx^2 + cx + d$ where $0 < b^2 < c$. Then $f(x)$:

Vedclass Test Series

Exam Paper Generator

Online Exam Module