To determine from which metals electrons will be emitted, we need to determine if the energy of the incident monochromatic light is greater than the work function of the metal. The energy of the incident light can be calculated using the formula:
E=‌where:
h is Planck's constant:
6.626×10−34‌Jsc is the speed of light:
3×108m∕ sλ is the wavelength of the light in meters
Given that the wavelength
λ=4770 (which is equivalent to
4770×10−10m ), we can substitute into the formula:
E=‌| 6.626×10−34×3×108 |
| 4770×10−10 |
By calculating this, we get:
E≈4.155×10−19JTo convert this energy from joules
(J) to electron volts
(eV), we use the conversion factor of
‌1‌eV=1.602×10−19J‌E≈‌| 4.155×10−19 |
| 1.602×10−19 |
‌eVThis simplifies to:
E≈2.59‌eVNow, we compare this energy with the work functions of the metals
A,B,C, and
D :
A: 4.2 eV
B: 3.7 eV
C: 3.2 eV
D: 2.3 eV
Since the energy of the incident light
(2.59‌eV) is less than the work functions of metals
A(4.2‌eV),
B(3.7‌eV), and
C(3.2‌eV), electrons will not be emitted from these metals. However, the energy is greater than the work function of metal
D(2.3‌eV), so electrons will be emitted from metal D .
Therefore, the correct answer is:
Option D: D only