Question

Calculate the maximum kinetic energy for emitted electron of Rubidium surface?

Short answer

Answer

Maximum kinetic energy of emitted electrons is $4.365\times {10}^{-19}\mathrm{J}$.

Step-by-step solution

Concept used

The formula to calculate frequency is as follows:

$\mathbf{\lambda }\mathbf{·}\mathbf{y}\mathbf{=}\mathbf{r}$

Where
$\mathbf{\lambda }$denotes wavelength.

v denotes frequency.

c denotes the speed of light.

As per Planck's hypothesis, the energy associated with photon is given as follow:

$\mathbf{E}\mathbf{=}\mathbf{h}\mathbf{·}\mathbf{v}$

Where,

v denotes frequency.

h denotes Planck's constant.

Avogadro's number is $\mathbf{6}\mathbf{.}\mathbf{0222}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}$atoms mol.

Calculate the Maximum Kinetic Energy

The conversion factor to convert nm to is as follows:

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

Thus, $254\mathrm{nm}$is converted to as follows:

$$\begin{gathered} \mathrm{Wavelength}=\left(254\mathrm{nm}\right)\left(\frac{{10}^{-9}\mathrm{m}}{1\mathrm{mm}}\right) \\ =2.54\times {10}^{-7}\mathrm{m} \end{gathered}$$

The conversion factor to convert J to is as follows:

$1\mathrm{J}={10}^{-3}\mathrm{kJ}$

Thus, $208.4\mathrm{kJ}/\mathrm{mol}$is converted to as follows:

$$\begin{gathered} \mathrm{Energy}=\left(208.4\mathrm{kJ}\right)\left(\frac{11}{{10}^{-3}\mathrm{kJ}}\right) \\ =2.084\times {10}^5\mathrm{J} \end{gathered}$$

Since $2.084\times {10}^5\mathrm{J}$is present for one mole that is, $6.022\times {10}^{23}$atoms

so energy required to ionize one rubidium atom is calculated as follows:

$$\begin{gathered} \mathrm{Energy}\mathrm{per}\mathrm{atom}=\left(1\mathrm{atoms}\right)\left(\frac{208\times 4\times {10}^3}{6002\times {10}^{33}}\right) \\ =3.46\times {10}^{-19}\mathrm{J}/\mathrm{atom} \end{gathered}$$

Thus to ionize one rubidium atom the energy required is $3.46\times {10}^{-10}\mathrm{J}$.

The formula to calculate frequency is as follows:

$\mathrm{v}=\frac{\mathrm{c}}{2}$

Value of c is $3\times {10}^8\mathrm{m}/\mathrm{s}$.

Value of $\mathrm{\lambda }$is $2.54\times {10}^{-7}m$.

Substitute the value in above equation

$$\begin{gathered} \mathrm{v}=\frac{\mathrm{s}}{2} \\ =\frac{3\times {10}^4\mathrm{ms}}{254\times {10}^{-7}\mathrm{m}} \\ =1.1811\times {10}^{15}{\mathrm{s}}^{-1} \end{gathered}$$

As per Planck's hypothesis, the energy associated with photon is given as follow:

$\mathrm{E}=\mathrm{h}-\mathrm{v}$

Value of v is $1.1811\times {10}^{15}{\mathrm{s}}^{-1}$.

Value of h is $6.626\times {10}^{-34}\mathrm{Js}$.

Substitute the value in above equation to calculate energy for single photon.

$$\begin{gathered} \mathrm{E}=\mathrm{h}·\mathrm{v} \\ =\left(6.626\times {10}^{-34}\mathrm{Js}\right)\left(1.1811\times {10}^{15}{\mathrm{s}}^{-1}\right) \\ =7.825\times {10}^{-19}\mathrm{J}/\mathrm{atom} \end{gathered}$$

Thus, maximum kinetic energy of emitted electrons can be calculated from difference in energies of radiation imparted and work function as follows:

$$\begin{gathered} \mathrm{Kinetic}\mathrm{energy}=\left(7.825\times {10}^{-19}-3.46\times {10}^{-19}\right)\mathrm{J} \\ =4.365\times {10}^{-19}\mathrm{J} \end{gathered}$$

Thus, maximum kinetic energy of emitted electrons is $4.365\times {10}^{-19}\mathrm{J}$.

Conclusion

Maximum kinetic energy of emitted electrons is $4.365\times {10}^{-19}\mathrm{J}$