$K^{\prime}=e V=50 \quad e V$
Total kinetic energy is equal to $150 \times 1.6 \times 10^{-19} J$
The de broglie wavelength is $\lambda=\frac{h}{\sqrt{2 m K}}=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-34} \times 150 \times 1.6 \times 10^{-19}}}=1 \times 10^{-10} \mathrm{m}$
$\lambda=1 A^{o}$
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$(i)$ $6.563 \times {10^{ - 19}}C$ $ (ii)$ $8.204 \times {10^{ - 19}}C$
$(iii)$ $11.50 \times {10^{ - 19}}C$ $ (iv)$ $13.13 \times {10^{ - 19}}C$
$(v)$ $16.48 \times {10^{ - 19}}C$ $ (vi)$ $18.09 \times {10^{ - 19}}C$
From this data the value of the elementary charge $(e)$ was found to be
