Question

The work functions for metals A,B and C are respectively 1.92 eV, 2.0 eV and 5 eV. According to Einstein's equation, the metals which will emit photoelectrons for a radiation of wavelength 4100 Å is/are:

• A. none
• B. A only
• C. A and B only
• D. all the three metals

Answer 1

The Correct Option is C

To determine which metals will emit photoelectrons when exposed to a radiation of wavelength 4100 Å, we need to use the photoelectric effect equation derived from Einstein's equation:

E = \dfrac{hc}{\lambda}

Where:

• E is the energy of the incident photon
• h is Planck's constant (6.626 \times 10^{-34} \, \text{Js})
• c is the speed of light (3 \times 10^8 \, \text{m/s})
• \lambda is the wavelength of the radiation (4100 Å = 4100 \times 10^{-10} \, \text{m})

First, let's calculate the energy of the incident photons:

E = \dfrac{6.626 \times 10^{-34} \times 3 \times 10^8}{4100 \times 10^{-10}}

Solving, we get:

E = \dfrac{1.9878 \times 10^{-25}}{4.1 \times 10^{-7}} = 4.848 \times 10^{-19} \, \text{J}

Converting this energy into electron volts (1 eV = 1.602 \times 10^{-19} \, \text{J}):

E = \dfrac{4.848 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 3.03 \, \text{eV}

Now, compare this energy with the work functions of the metals:

• Metal A: 1.92 eV
• Metal B: 2.0 eV
• Metal C: 5 eV

The energy of the incident photon (3.03 eV) is greater than the work functions of metals A and B. Thus, both of these metals will emit photoelectrons.

The energy is not greater than the work function for Metal C, so no photoelectrons will be emitted from Metal C.

Therefore, the correct answer is: A and B only.