Question

Find the accelerating potential (V) that must be imparted to a helium atom so that its wavelength is 5 A${}^o$(1 amu = 1.67 $\times$ 10${}^{-24}$ g)

• A. $8.2\times$ 10${}^{-6}$
• B. $16.2\times$ 10${}^{-6}$
• C. $8.2\times$ 10${}^{-4}$
• D. None of these

Answer 1

The correct option is C $8.2\times$ 10${}^{-4}$

As we know,
$\lambda$ = $\frac{h}{p}$ = $\frac{h}{mv}$,
E = $\frac{1}{2}$ mv${}^2$
mv${}^2$ = 2E
m $\times$ mv${}^2$ = 2E $\times$ m
mv = $\sqrt{2Em}$
p =$\sqrt{2Em}$]
$\lambda$ = h/P = $\frac{h}{\sqrt{2EM}}$
m(He) = 4 $\times$ 1.67 $\times$ 10${}^{-27}$ kg
Therefore, substituting value in equation
5 $\times$ 10${}^{-10}$ m = $\frac{6.62\times {10}^{-34}}{\sqrt{2}.\sqrt{E}\times (4\times 1.67\times {10}^{-27}{)}^{\frac{1}{2}}}$
$\therefore$ E = $\frac{(6.62\times {10}^{-34}{)}^{\frac{1}{2}}}{\sqrt{2}.\sqrt{E}\times (4\times 1.67\times {10}^{-27}{)}^{\frac{1}{2}}}$

=$\frac{43.8244\times {10}^{-68}}{334\times {10}^{-27}}$ = 0.1312 $\times$ 10${}^{-21}$

E = eV

V = $\frac{E}{e}$ = $\frac{0.1312\times {10}^{-21}}{1.6\times {10}^{-19}}$

$\therefore V=8.2\times {10}^{-4}$