A photon of wavelength 4 x 10⁻⁷ m strikes a metal surface, and the work function of the metal is 2.1

Q: What is the correct answer for CBSE11?

The correct answer is A. First, calculate the energy of the incident photon (E_photon): E_photon = hc/λ = (6.626 x 10⁻³⁴ J s x 3.00 x 10⁸ m s⁻¹) / (4 x 10⁻⁷ m) = 4.9695 x 10⁻¹⁹ J. Next, convert the work function (W₀) from eV to Joules: W₀ = 2.13 eV x 1.602 x 10⁻¹⁹ J/eV = 3.41226 x 10⁻¹⁹ J. Finally, calculate the kinetic energy (KE) of the emitted electron using Einstein's equation: KE = E_photon - W₀ = (4.9695 x 10⁻¹⁹ J)

Q: A photon of wavelength 4 x 10⁻⁷ m strikes a metal surface, and the work function of the metal is 2.13 eV. Calculate the kinetic energy of the emitted electron in Joules. (1 eV = 1.602 x 10⁻¹⁹ J, h =…

The correct answer is A. First, calculate the energy of the incident photon (E_photon): E_photon = hc/λ = (6.626 x 10⁻³⁴ J s x 3.00 x 10⁸ m s⁻¹) / (4 x 10⁻⁷ m) = 4.9695 x 10⁻¹⁹ J. Next, convert the work function (W₀) from eV to Joules: W₀ = 2.13 eV x 1.602 x 10⁻¹⁹ J/eV = 3.41226 x 10⁻¹⁹ J. Finally, calculate the kinetic energy (KE) of the emitted electron using Einstein's equation: KE = E_photon - W₀ = (4.9695 x 10⁻¹⁹ J)

Q: What are the options for this CBSE11 question?

A. 1.85 x 10⁻¹⁹ J · B. 4.97 x 10⁻¹⁹ J · C. 3.41 x 10⁻¹⁹ J · D. 1.56 x 10⁻¹⁹ J

Question

A photon of wavelength 4 x 10⁻⁷ m strikes a metal surface, and the work function of the metal is 2.13 eV. Calculate the kinetic energy of the emitted electron in Joules. (1 eV = 1.602 x 10⁻¹⁹ J, h = 6.626 x 10⁻³⁴ J s, c = 3.00 x 10⁸ m s⁻¹)

FirstInTest · Accepted Answer

First, calculate the energy of the incident photon (E_photon): E_photon = hc/λ = (6.626 x 10⁻³⁴ J s x 3.00 x 10⁸ m s⁻¹) / (4 x 10⁻⁷ m) = 4.9695 x 10⁻¹⁹ J. Next, convert the work function (W₀) from eV to Joules: W₀ = 2.13 eV x 1.602 x 10⁻¹⁹ J/eV = 3.41226 x 10⁻¹⁹ J. Finally, calculate the kinetic energy (KE) of the emitted electron using Einstein's equation: KE = E_photon - W₀ = (4.9695 x 10⁻¹⁹ J) - (3.41226 x 10⁻¹⁹ J) = 1.55724 x 10⁻¹⁹ J. The closest option is 1.85 x 10⁻¹⁹ J. Let's recheck the calculation. The closest option is 1.56 x 10⁻¹⁹ J. Let's re-evaluate the options. The provided solution in the textbook for a similar problem (2.9) gives 1.988 x 10⁻¹⁹ J. Let's recheck the values. E_photon = 4.97 x 10⁻¹⁹ J. W₀ = 3.41 x 10⁻¹⁹ J. KE = (4.97 - 3.41) x 10⁻¹⁹ J = 1.56 x 10⁻¹⁹ J. So, the correct option is D.

Answer

1.85 x 10⁻¹⁹ J

Answer

4.97 x 10⁻¹⁹ J

Answer

3.41 x 10⁻¹⁹ J

Answer

1.56 x 10⁻¹⁹ J

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