Chemistry - Atomic Structure Question with Solution | TestHub

${\lambda }^2=\frac{h^2}{2mqV}$, so the slope of ${\lambda }^2$ against $V^{-1}$ is $\frac{h^2}{2mq}=\tan \theta =0.9$.

$m=\frac{h^2}{2q(0.9)}=\frac{36\times 10^{-68}}{2\times 1.6\times 10^{-19}\times 0.9}=\frac{36\times 10^{-68}}{2.88\times 10^{-19}}=1.25\times 10^{-48}\mathrm{kg}$.

(1.39 comes from squaring the slope by mistake; the last option has the wrong unit.)

Answer: (A).