Chapter 26, Problem 63QAP

To determine

(a)

The de Broglie wavelength of electron of kinetic energy $1.60\times {10}^{-19}\text{J}$

Answer to Problem 63QAP

The de Broglie wavelength of the electron is $\text{1}\text{.23}\times {\text{10}}^{\text{-9}}\text{ m}$

Explanation of Solution

Given:

The kinetic energy of the electron is $K=1.60\times {10}^{-19}\text{J}$

Formula used:

The de Broglie wavelength expression is given by,

  $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mK}}$

Calculation:

Putting the values of h, m and K in the above equation we get

  $\lambda =\frac{6.63\times {10}^{-34}}{\sqrt{2\times 9.11\times {10}^{-31}\times 1.6\times {10}^{-19}}}\text{ m = 1}\text{.23}\times {\text{10}}^{\text{-9}}\text{ m}$

Conclusion:

So, the de Broglie wavelength of the electron is $\text{1}\text{.23}\times {\text{10}}^{\text{-9}}\text{ m}$

To determine

(b)

The de Broglie wavelength of electron of kinetic energy $1.60\times {10}^{-18}\text{J}$

Answer to Problem 63QAP

The de Broglie wavelength of the electron is $\text{3}\text{.89}\times {\text{10}}^{\text{-10}}\text{ m}$

Explanation of Solution

Given:

The kinetic energy of the electron is $K=1.60\times {10}^{-18}\text{J}$.

Formula used:

The de Broglie wavelength expression is given by,

  $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mK}}$

Calculation:

Putting the values of h, m and K in the above equation we get

  $\lambda =\frac{6.63\times {10}^{-34}}{\sqrt{2\times 9.11\times {10}^{-31}\times 1.6\times {10}^{-18}}}\text{ m = 3}\text{.89}\times {\text{10}}^{\text{-10}}\text{ m}$

Conclusion:

So, the de Broglie wavelength of the electron is $\text{3}\text{.89}\times {\text{10}}^{\text{-10}}\text{ m}$

To determine

(c)

The de Broglie wavelength of electron of kinetic energy $1.60\times {10}^{-17}\text{J}$

Answer to Problem 63QAP

The de Broglie wavelength of the electron is $\text{1}\text{.22}\times {\text{10}}^{\text{-10}}\text{ m}$

Explanation of Solution

Given:

The kinetic energy of the electron is $K=1.60\times {10}^{-17}\text{J}$

Formula used:

The de Broglie wavelength expression is given by,

  $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mK}}$

Calculation:

Putting the values of h, m and K in the above equation we get,

  $\lambda =\frac{6.63\times {10}^{-34}}{\sqrt{2\times 9.11\times {10}^{-31}\times 1.6\times {10}^{-17}}}\text{ m = 1}\text{.22}\times {\text{10}}^{\text{-10}}\text{ m}$

Conclusion:

So, the de Broglie wavelength of the electron is $\text{1}\text{.22}\times {\text{10}}^{\text{-10}}\text{ m}$

To determine

(d)

The de Broglie wavelength of electron of kinetic energy $1.60\times {10}^{-16}\text{J}$

Answer to Problem 63QAP

The de Broglie wavelength of the electron is $\text{3}{\text{.86X10}}^{\text{-11}}\text{ meter}$

Explanation of Solution

Given:

The kinetic energy of the electron is $K=1.60\times {10}^{-16}\text{J}$

Formula used:

The de Broglie wavelength expression is given by

  $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mK}}$

Calculation:

Putting the values of h, m and K in the above equation we get

  $\lambda =\frac{6.63\times {10}^{-34}}{\sqrt{2\times 9.11\times {10}^{-31}\times 1.6\times {10}^{-16}}}\text{ m = 3}\text{.86}\times {\text{10}}^{\text{-11}}\text{ m}$

Conclusion:

So, the de Broglie wavelength of the electron is $\text{3}\text{.86}\times {\text{10}}^{\text{-11}}\text{ m}$