# In a sample of excited hydrogen atoms electrons make transition from n = 3 to n = 1 and emitted photons strike on a metal of work function 6.08 eV. The maximum kinetic energy of the ejected electron in the process will be?

## 4 eV 6 eV 2 eV 8 eV

Nov 20, 2017

$\approx$ $\text{6 eV}$

#### Explanation:

The idea here is that in order for an electron to be ejected from the surface of the metal, the energy of the incoming photon must exceed the work function of the metal.

So you can say that you have

color(blue)(ul(color(black)("K"_ "E" = E_"photon" - W)))" " " "color(darkorange)(("*"))

Here

• $\text{K"_"E}$ is the kinetic energy of the ejected electron
• ${E}_{\text{photon}}$ is the energy of the photon
• $W$ is the work function of the metal

So in order to have $\text{K"_"E} > 0$ you need to have

${E}_{\text{photon}} > W$

The kinetic energy of the electron will reach its maximum value when the electron absorbs all the energy of the photon that does not go into overcoming the work function.

So use the Rydberg equation and the Planck - Einstein relation to find the energy of the photons emitted during the ${n}_{i} = 3 \to {n}_{f} = 1$ transition, which is part of the Lyman series.

You have

$\frac{1}{l a m \mathrm{da}} = R \cdot \left(\frac{1}{n} _ {f}^{2} - \frac{1}{n} _ {i}^{2}\right)$

Here

• $R$ is the Rydberg constant, equal to $1.097 \cdot {10}^{7}$ ${\text{m}}^{- 1}$
• $l a m \mathrm{da}$ is the wavelength of the photon

and

$E = \frac{h \cdot c}{l} a m \mathrm{da}$

Here

• $E$ is the energy of the photon
• $h$ is Planck's constant, equal to $6.626 \cdot {10}^{- 34} \textcolor{w h i t e}{.} \text{J s}$
• $c$ is the speed of light in a vacuum, usually given as $3 \cdot {10}^{8} \textcolor{w h i t e}{.} {\text{m s}}^{- 1}$

You can combine these two equations to get

$\frac{h \cdot c}{E} = \frac{1}{R} \cdot \frac{1}{\left(\frac{1}{n} _ {f}^{2} - \frac{1}{n} _ {i}^{2}\right)}$

Rearrange to solve for $E$

$E = \frac{{n}_{i}^{2} - {n}_{f}^{2}}{{n}_{i}^{2} \cdot {n}_{f}^{2}} \cdot h \cdot R \cdot c$

Plug in the values to find

E = (3^2 - 1^2)/(3^2 * 1^2) * 6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * 1.097 * 10^7color(red)(cancel(color(black)("m"^(-1)))) * 3 * 10^8 color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1))))

$E = 1.938 \cdot {10}^{- 18}$ $\text{J}$

To convert this to electronvolts, use the fact that

$\text{1 eV} = 1.6 \cdot {10}^{- 19}$ $\text{J}$

You will end up with

1.938 * 10^(-18) color(red)(cancel(color(black)("J"))) * "1 eV"/(1.6 * 10^(-19)color(red)(cancel(color(black)("J")))) = "12.11 eV"

Finally, use equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\left(\text{*}\right)}$ to find the maximum kinetic energy of the ejected electron.

$\text{K"_"E" = "12.11 eV" - "6.08 eV}$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{K"_"E" = "6.03 eV" ~~ "6 eV}}}}$

So I would say that your answer is (2) $\text{6 eV}$.