The work functions of cesium (Cs) and lithium (Li) metals are 1.9 eV and 2.5 eV , respectively. If we incident a light of wavelength 550 nm on these two metal surfaces, then photo-electric effect is possible for the case of

Question

Aicharya · Accepted Answer

Step 1: Calculate the Energy of the Incident Photon
The energy of a photon is given by the equation:
$E=\frac{h c}{\lambda}$where:
$E=\frac{h c}{\lambda}$ where: $\begin{aligned} & h=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{~s} \text { (Planck's constant), } \\ & c=3.0 \times 10^{8} \mathrm{~m} / \mathrm{s} \text { (speed of light), } \\ & \lambda=550 \mathrm{~nm}=550 \times 10^{-9} \mathrm{~m} \end{aligned}$ First, calculate the photon energy in joules:
$\begin{gathered}
E=\frac{\left(6.626 \times 10^{-34}\right)\left(3.0 \times 10^8\right)}{550 \times 10^{-9}} \\
E=\frac{1.9878 \times 10^{-25}}{550 \times 10^{-9}} \\
E=3.615 \times 10^{-19} \mathrm{~J}
\end{gathered}$
Convert this to electron volts (eV) using $1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}$ :
$\begin{gathered}
E=\frac{3.615 \times 10^{-19}}{1.6 \times 10^{-19}} \\
E \approx 2.26 \mathrm{eV}
\end{gathered}$
Step 2: Compare Photon Energy with Work Functions
Cesium ($\phi_{\mathrm{Cs}}=1.9 \mathrm{eV}$)
Since $E_{\text {photon }}=2.26 \mathrm{eV}$ is greater than $\phi_{\mathrm{Cs}}=1.9 \mathrm{eV}$,photoelectric emission occurs.
Lithium ($\phi_{\mathrm{Li}}=2.5 \mathrm{eV}$)
Since $E_{\text {photon }}=2.26 \mathrm{eV}$ is less than $\phi_{\mathrm{Li}}=2.5 \mathrm{eV}$, photoelectric emission does not occur.
Conclusion:
Photoelectric effect is possible only for Cesium (Cs), but not for Lithium (Li).

The work functions of cesium (Cs) and lithium (Li) metals are 1.9 eV and 2.5 eV , respectively. If we…

Solution