Question

The work function of a metal M is 6.3 eV. The wave length of the incident radiation required to just eject the electrons from its surface (in nm) is

• A. 102
• B. 330
• C. 197
• D. 310

Hint:

Threshold condition for photoelectric effect: Photon energy $E_{photon} = W_o$.
Photon energy $E_{photon} = hc/\lambda$.
Threshold wavelength $\lambda_{threshold} = hc/W_o$.
Use $hc \approx 1240 \text{ eV nm}$ or $1242 \text{ eV nm}$ when $W_o$ is in eV and $\lambda$ is needed in nm.
Ensure $W_o$ is in eV for this shortcut. If $W_o$ is in Joules, use $h$ and $c$ in SI units to get $\lambda$ in meters.

Answer 1

The Correct Option is C

To "just eject" electrons, the energy of the incident photons ($E_{photon}$) must be equal to the work function ($W_o$) of the metal. This corresponds to the threshold condition. $E_{photon} = W_o$. The energy of a photon is given by $E_{photon} = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength. So, $\frac{hc}{\lambda} = W_o$. This gives the threshold wavelength $\lambda_{threshold} = \frac{hc}{W_o}$. Given values: Work function $W_o = 6.3 \text{ eV}$. We need to find $\lambda$ in nm. A useful formula for this conversion is $\lambda (\text{nm}) = \frac{hc (\text{eV nm})}{W_o (\text{eV})}$. The value of $hc$ is approximately $1240 \text{ eV nm}$ or $1242 \text{ eV nm}$. Let's use $1240 \text{ eV nm}$. $\lambda = \frac{1240 \text{ eV nm}}{6.3 \text{ eV}}$. $\lambda = \frac{1240}{6.3} \text{ nm}$. Calculation: $\frac{1240}{6.3} = \frac{12400}{63}$. $12400 \div 63$: $12400 / 63 \approx 196.825...$ So, $\lambda \approx 196.8 \text{ nm}$. Let's try with $hc \approx 1242 \text{ eV nm}$: $\lambda = \frac{1242}{6.3} \text{ nm} = \frac{12420}{63} \text{ nm}$. $12420 \div 63 \approx 197.14...$ So, $\lambda \approx 197.1 \text{ nm}$. Comparing with the options: % Option (a) 102 nm % Option (b) 330 nm % Option (c) 197 nm % Option (d) 310 nm The calculated value $\lambda \approx 197.1 \text{ nm}$ is closest to option (c) 197 nm. This suggests $hc \approx 1242 \text{ eV nm}$ or a similar value was used. (If using $h=6.626 \times 10^{-34}$ Js, $c=2.998 \times 10^8$ m/s, $e=1.602 \times 10^{-19}$ C, then $hc/e \approx 1239.8$ eV nm). $1239.8 / 6.3 \approx 196.8$ nm. If $h=6.63 \times 10^{-34}$ Js: $hc/e \approx 1243.1$ eV nm. $1243.1 / 6.3 \approx 197.3$ nm. Option (c) 197 nm is the best fit. \[ \boxed{197 \text{ nm}} \]