Question

Calculate the maximum wavelength that can release an electron from a lithium atom.

Short answer

Answer

The maximum wavelength that can release an electron from a lithium atom is 428nm.

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 a photon is given as follows:

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

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 wavelength

The wavelength of absorption by C atom 150nm.

The work function for lithium is 279.7kJ/mol.

The conversion factor to convert J to kJ is as follows:

$\mathrm{lJ}={10}^{-3}\mathrm{kJ}$

Thus, 279.7kJ/mol is converted to as follows:

$$\begin{gathered} \mathrm{Energy}=\left(279.7\mathrm{kJ}\right)\left(\frac{1\mathrm{J}}{{10}^{-1}\mathrm{k}}\right) \\ =2.797\times {10}^5\mathrm{J} \end{gathered}$$

The formula to calculate frequency is as follows:

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

When $\mathrm{\lambda }$is $1.5\times {10}^{-7}\mathrm{m}$.

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

Substitute the value in the above equation.

$$\begin{gathered} \mathrm{y}=\frac{\mathrm{c}}{2} \\ =\frac{3\times {10}^4\mathrm{ms}}{1.5\times {10}^{-7}\mathrm{m}} \\ =2\times {10}^{15}{\mathrm{s}}^{-1} \end{gathered}$$

Since $2.797\times {10}^5\mathrm{J}$is present for one mole that is, $6.022\times {10}^{23}$atoms so energy per atom is calculated as follows:

$$\begin{gathered} \mathrm{Energy}\mathrm{per}\mathrm{atom}=\left(1\mathrm{atom}\right)\left(\frac{2707\times {10}^3\mathrm{J}}{6.002\times {10}^{23}\mathrm{atoms}}\right) \\ =4.644\times {10}^{-19}\mathrm{J}/\mathrm{atom} \end{gathered}$$

As per Einstein's photoelectric theory, the minimum energy required to release an electron is given as follows:

${\mathrm{E}}_0=\mathrm{h}·\frac{\mathrm{c}}{{\mathrm{a}}_0}$

Rearrange to obtain expression of ${\mathrm{\lambda }}_0$.

${\mathrm{\lambda }}_0=\mathrm{h}·\frac{\in }{{\mathrm{h}}_0}$

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

The value of E₀ is $4.644\times {10}^{-19}\mathrm{J}$.

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

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

$$\begin{gathered} {\lambda }_0=h-\frac{c}{5_0} \\ =\frac{\left(i.266\times {10}^{34}Js\right)\left(3\times {10}^{\times }m/s\right)}{\left(4.644\times {10}^{-9}J\right)} \\ =4.28\times {10}^{-7}\mathrm{m} \end{gathered}$$

The conversion factor to convert to is as follows:

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

Thus, $4.28\times {10}^{-7}\mathrm{m}$is converted to as follows:

$\mathrm{Wavelength}=\left(4.28\times {10}^{-7}\mathrm{m}\right)\left(\frac{1\mathrm{nm}}{{10}^{-9}\mathrm{m}}\right)\approx 428\mathrm{nm}$

Thus, a maximum wavelength that can release an electron from a lithium atom is 428nm.

Conclusion

The maximum wavelength that can release an electron from a lithium atom is 428nm.