Question

Work function (W$_o$) values of six metals (in eV) are given below: Li \quad Mg \quad Cu \quad Ag \quad K \quad Na 2.42 \quad 3.7 \quad 4.8 \quad 4.3 \quad 2.25 \quad 2.3 How many of the above metals do not eject the electrons when they are made to strike with radiation of wavelength 400 nm? (h=$6.62 \times 10^{-34}$ Js)

• A. 4
• B. 2
• C. 3
• D. 5

Hint:

Condition for photoelectric effect: Photon energy $E_{photon} \ge W_o$ (work function).
Photon energy $E_{photon} = hc/\lambda$.
A useful approximation for photon energy: $E (\text{eV}) \approx \frac{1240}{\lambda (\text{nm})}$.
Compare the calculated photon energy with each metal's work function.

Answer 1

The Correct Option is C

For electrons to be ejected from a metal surface (photoelectric effect), the energy of the incident photons ($E_{photon}$) must be greater than or equal to the work function ($W_o$) of the metal. $E_{photon} \ge W_o$. If $E_{photon}<W_o$, electrons are not ejected. First, calculate the energy of the incident photons. The energy of a photon is given by $E_{photon} = hf = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the radiation. Given values: Wavelength $\lambda = 400 \text{ nm} = 400 \times 10^{-9} \text{ m} = 4 \times 10^{-7} \text{ m}$. Planck's constant $h = 6.62 \times 10^{-34} \text{ Js}$. (The image shows $6.62 \times 10^{-34}$ Js, typically $6.626 \times 10^{-34}$ Js is used). Speed of light $c \approx 3 \times 10^8 \text{ m/s}$. Calculate $E_{photon}$ in Joules: $E_{photon} = \frac{(6.62 \times 10^{-34} \text{ Js}) \times (3 \times 10^8 \text{ m/s})}{4 \times 10^{-7} \text{ m}}$. $E_{photon} = \frac{6.62 \times 3}{4} \times \frac{10^{-34} \times 10^8}{10^{-7}} \text{ J}$. $\frac{6.62 \times 3}{4} = \frac{19.86}{4} = 4.965$. $\frac{10^{-26}}{10^{-7}} = 10^{-26 - (-7)} = 10^{-26 + 7} = 10^{-19}$. So, $E_{photon} = 4.965 \times 10^{-19} \text{ J}$. Convert $E_{photon}$ to electron-volts (eV), since work functions are given in eV. $1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}$. (Using $1.6 \times 10^{-19}$ J for simplicity if not specified). $E_{photon} (\text{in eV}) = \frac{4.965 \times 10^{-19} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}}$. (Using $1.602$ for better precision). $E_{photon} (\text{in eV}) \approx \frac{4.965}{1.602} \approx 3.10 \text{ eV}$. A common shortcut for photon energy in eV when wavelength is in nm is $E(eV) \approx \frac{1240}{\lambda(nm)}$ or $\frac{1242}{\lambda(nm)}$. Using $\lambda = 400 \text{ nm}$: $E_{photon} \approx \frac{1240 \text{ eV nm}}{400 \text{ nm}} = \frac{124}{40} = \frac{31}{10} = 3.1 \text{ eV}$. Or using $hc = (6.626 \times 10^{-34}) \times (2.998 \times 10^8) \approx 1.986 \times 10^{-25} \text{ Jm}$. $1.986 \times 10^{-25} \text{ Jm} / (1.602 \times 10^{-19} \text{ J/eV}) \approx 1239.8 \text{ eV nm}$. So 1240 is good. $E_{photon} \approx 3.10 \text{ eV}$. Now compare $E_{photon} \approx 3.10 \text{ eV}$ with the work functions ($W_o$) of the metals: \begin{itemize} \item Li: $W_o = 2.42 \text{ eV}$. Since $3.10>2.42$, electrons are ejected. \item Mg: $W_o = 3.7 \text{ eV}$. Since $3.10<3.7$, electrons are NOT ejected. \item Cu: $W_o = 4.8 \text{ eV}$. Since $3.10<4.8$, electrons are NOT ejected. \item Ag: $W_o = 4.3 \text{ eV}$. Since $3.10<4.3$, electrons are NOT ejected. \item K: $W_o = 2.25 \text{ eV}$. Since $3.10>2.25$, electrons are ejected. \item Na: $W_o = 2.3 \text{ eV}$. Since $3.10>2.3$, electrons are ejected. \end{itemize} The metals that do NOT eject electrons are Mg, Cu, and Ag. There are 3 such metals. This matches option (c). \[ \boxed{3} \]