Work function of a metal is 2.1eV. Which of the waves of the following wavelengths will be able to emit photoelectrons from its surfacea)4000 Å, 7500 Åb)5500 Å, 6000 Åc)4000 Å, 6000 Åd)None of theseCorrect answer is option 'D'. Can you explain this answer?

Question

Work function of a metal is 2.1eV. Which of the waves of the following wavelengths will be able to emit photoelectrons from its surfacea)4000 &#8491;, 7500 &#8491;b)5500 &#8491;, 6000 &#8491;c)4000 &#8491;, 6000 &#8491;d)None of theseCorrect answer is option 'D'. Can you explain this answer?

Answer

First u find threshold wavelenth it comes out to be 5904 A .This is the maximum wavelength .above it not possible ok.

Answer

Work function is 2.1ev and for emmiting photoelectrons from metal surface it's energy is higher than the 2.1ev , for 4000A′ energy is 12400/4000= 3.1ev so it's ok but for 6000A' energy is 12400/6000=2.06ev  is less than work function of metal and for 7000A energy is 12400/7000= 1.77,ev  and for 5500 energy is 12400/5500=2.25 ev. so none of the option is correct for that then option D is correct

Answer

Understanding Work Function
The work function (Φ) of a metal is the minimum energy required to remove an electron from its surface. In this case, the given work function is 2.1 eV.
Energy of Photons
The energy of a photon can be calculated using the formula:
Energy (E) = hc/λ
where:
- h = Planck's constant (4.14 x 10^-15 eV·s)
- c = speed of light (3 x 10^8 m/s)
- λ = wavelength in meters
Converting Wavelengths to Energy
To determine if the given wavelengths can emit photoelectrons, we need to convert them to energy and compare with the work function.
Wavelengths in the Options
- a) 4000 Å (400 nm) and 7500 Å (750 nm)
- b) 5500 Å (550 nm) and 6000 Å (600 nm)
- c) 4000 Å (400 nm) and 6000 Å (600 nm)
Calculating Energies
1. 4000 Å (400 nm):
E = hc/λ = (4.14 x 10^-15 eV·s * 3 x 10^8 m/s) / (400 x 10^-9 m) ≈ 3.1 eV.
2. 7500 Å (750 nm):
E = (4.14 x 10^-15 eV·s * 3 x 10^8 m/s) / (750 x 10^-9 m) ≈ 1.65 eV.
3. 5500 Å (550 nm):
E ≈ 2.25 eV.
4. 6000 Å (600 nm):
E ≈ 2.07 eV.
Comparison with Work Function
- 4000 Å: 3.1 eV (Greater than 2.1 eV) - Can emit electrons.
- 7500 Å: 1.65 eV (Less than 2.1 eV) - Cannot emit electrons.
- 5500 Å: 2.25 eV (Greater than 2.1 eV) - Can emit electrons.
- 6000 Å: 2.07 eV (Greater than 2.1 eV) - Can emit electrons.
Conclusion
None of the pairs of wavelengths in the options consistently emit photoelectrons based on the work function of 2.1 eV. Therefore, the correct answer is option D: None of these.

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