NEET Questions Solved

NEET - 2016

Electrons of mass m with de-Brogile wavelength $\lambda$ fall on the target in an X-ray tube. The cut off wavelength $\left({\lambda }_{0}\right)$ of the emitted X-ray is

(a) ${\lambda }_{0}=\frac{2mc{\lambda }^{2}}{h}$          (b) ${\lambda }_{0}=\frac{2h}{mc}$

(c) ${\lambda }_{0}=\frac{2{m}^{2}{c}^{2}{\lambda }^{3}}{{h}^{2}}$       (d) ${\lambda }_{0}=\lambda$

(a) Key idea Cut-off wavelength occurs when incoming electron looses its complete energy in collision. This energy appears in the form of X-rays.

Given, mass of electrons=m

de-Broglie wavelength=$\lambda$

So, kinet energy, of  electron = $\frac{{p}^{2}}{2m}$

$={\left(\frac{h}{\lambda }\right)}^{2}=\frac{{h}^{2}}{2m{\lambda }^{2}}$

Now, maximum energy of photon can be given by

Difficulty Level:

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