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Q.
The accelerating potential (in volt) that must be imparted
to a proton beam to give it an effective wavelength of
$0.005 \,nm$ is________ .
Structure of Atom
Solution:
$u =\frac{ h }{ m \lambda}$
$\therefore $ mass of proton $=1.67 \times 10^{-27} kg $
$u =\frac{6.625 \times 10^{-34}}{1.67 \times 10^{-27} \times 0.005 \times 10^{-9}}$
$=7.94 \times 10^{4} $ metre $ \sec ^{-1}$
Now accelerating potential is $V$, then velocity $(u)$ acquired by the charge particle having charge $Q$ and mass $m$.
$\because Q . V =\frac{1}{2} mu ^{2}$
$u =\sqrt{\left(\frac{2 QV }{ m }\right)}=\sqrt{\left(\frac{2 \times 1.602 \times 10^{-9} \times V }{1.67 \times 10^{-27}}\right)}$
or $7.94 \times 10^{4}=\sqrt{\left(\frac{2 \times 1.602 \times 10^{-19} \times V }{1.67 \times 10^{-27}}\right)}$
$V =32.85$ volt