Question

2. Calculate the velocity and the kinetic energy of an electron whose wave
length is 896.7 nm. (Write the final answer only. You need not write the
steps involved here). *​

Answer 1

Answer:

$\bold{\sf{v=7\times{10}^{5}\;m{s}^{-1}}}$.

$\bold{\sf{K.E = 2.22\times{10}^{-19}\;J}}$

Explanation:

Given that,

Wavelength of an electron is = 896.7 nm

But, we know that,

$1 \: nm = {10}^{ - 9} \: m$

Therefore, we have,

$= > \lambda = 896.7 \: \times {10}^{ - 9} \: m \\ \\ = > \lambda = 8.967 \times {10}^{ - 7} \: m$

To find it's Kinetic Energy.

We know that,

Kinetic Energy = $\dfrac{hc}{\lambda}$

Where,

• $h=6.626\times{10}^{-34}\;J\:s$
• $c= 3 \times {10}^{8}\; m{s}^{-1}$

Therefore, we will get,

$= > K.E = \frac{hc}{\lambda}$

Substituting the values, we get,

$= > K.E = \frac{6.626 \times {10}^{ - 34 } \times 3 \times {10}^{8} }{8.967 \times {10}^{ - 7} } \\ \\ = > K.E = 2.22 \times {10}^{ - 34 + 8 + 7} \\ \\ = > K.E = 2.22 \times {10}^{ - 34 + 15} \\ \\ = > K.E = 2.22 \times {10}^{ - 19}$

Therefore, Kinetic Energy is $\bold{2.22\times{10}^{-19}\;J}$

Now, we know that,

Kinetic Energy = $\dfrac{1}{2}m{v}^{2}$.

Where,

• m is mass of the electron
• v is the required velocity.

Also, we know that,

Mass of an electron, m = $9.1\times{10}^{-31}\;Kg$

Therefore, we will get,

$= > \frac{1}{2} \times 9.1 \times {10}^{ - 31} \times {v}^{2} = 2.22 \times {10}^{ - 19} \\ \\ = > {v}^{2} = \frac{2 \times 2.22 \times {10}^{ - 19} }{9 .1 \times {10}^{ - 31} } \\ \\ = > {v}^{2} = \frac{4.44 \times {10}^{ - 19 + 31} }{9.1} \\ \\ = > {v}^{2} = 0.49 \times {10}^{12} \\ \\ = > v = \sqrt{0.49 \times {10}^{12} } \\ \\ = > v = 0.7 \times {10}^{6} \\ \\ = > v = 7 \times {10}^{5}$

Hence, required velocity of electron is $\bold{7\times{10}^{5}\;m{s}^{-1}}$.