The mass of an electron is $9.1 \times 10^{-31} \ kg$. If its $K.E.$ is $3.0 \times 10^{-25} \ J$,calculate its wavelength.
The kinetic energy $(K.E.)$ is given by the formula: $K.E. = \frac{1}{2}mv^2$.
First,calculate the velocity $(v)$:
$v = \sqrt{\frac{2 \times K.E.}{m}} = \sqrt{\frac{2 \times 3.0 \times 10^{-25} \ J}{9.1 \times 10^{-31} \ kg}} \approx 812 \ m \ s^{-1}$.
Now,use the de Broglie wavelength formula: $\lambda = \frac{h}{mv}$.
Substituting the values: $\lambda = \frac{6.626 \times 10^{-34} \ J \ s}{(9.1 \times 10^{-31} \ kg) \times (812 \ m \ s^{-1})}$.
$\lambda \approx 8.967 \times 10^{-7} \ m = 896.7 \ nm$.
First,calculate the velocity $(v)$:
$v = \sqrt{\frac{2 \times K.E.}{m}} = \sqrt{\frac{2 \times 3.0 \times 10^{-25} \ J}{9.1 \times 10^{-31} \ kg}} \approx 812 \ m \ s^{-1}$.
Now,use the de Broglie wavelength formula: $\lambda = \frac{h}{mv}$.
Substituting the values: $\lambda = \frac{6.626 \times 10^{-34} \ J \ s}{(9.1 \times 10^{-31} \ kg) \times (812 \ m \ s^{-1})}$.
$\lambda \approx 8.967 \times 10^{-7} \ m = 896.7 \ nm$.
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