IIT-JEE 2007 · Paper 2 · Physics Q9

PhysicsSingle correct+3 / -1Question 9 of 66

Electrons with de-Broglie wavelength $\lambda$ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is

\lambda_0=\dfrac{2mc\lambda^2}{h}

\lambda_0=\dfrac{2h}{mc}

\lambda_0=\dfrac{2m^2c^2\lambda^3}{h^2}

\lambda_0=\lambda

Answer and worked solution

Official answer: A

Electron energy. $p=\dfrac h\lambda$ , so $K=\dfrac{p^2}{2m}=\dfrac{h^2}{2m\lambda^2}$ .

Cut-off. The shortest X-ray wavelength takes all of this energy: $\dfrac{hc}{\lambda_0}=K$ ,
$\lambda_0=\frac{hc}{K}=\frac{2mc\lambda^2}h.$

(A)

Topics:X-rays de Broglie waves

IIT-JEE 2007 · Paper 2, Physics question 9