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Chapter 17: Problem 2387

Work functions for tungsten and sodium are \(4.5 \mathrm{eV}\) and $2.3 \mathrm{eV}\( respectively. If threshold wavelength of sodium is \)5460 \AA$, threshold wavelength for tungsten will be \(\ldots \ldots . . \AA\) (A) 528 (B) 10683 (C) 2791 (D) 5893

Short Answer

Expert verified
The threshold wavelength for tungsten is approximately 3102 Å, which is close to option (C) 2791 Å. This is not an exact match, so keep in mind that there may be some rounding error in the provided options.

Step by step solution

01

Recall the work function equation

The work function (Φ) is the minimum energy required to remove an electron from the surface of a metal. The work function equation is as follows: Φ = h*c/λ Where, Φ = work function (in eV) h = Planck's constant (6.626 x 10^{-34} Js) c = speed of light (3 x 10^8 m/s) λ = threshold wavelength (in meters)
02

Convert the threshold wavelength of sodium to meters

Given the threshold wavelength of sodium is 5460 Å, we need to convert it into meters: 1 Å = 10^{-10} meters So, 5460 Å = 5460 * (10^{-10}) meters = 5.460 x 10^{-7} meters
03

Calculate the work function of sodium in Joules

The work function of sodium is given as 2.3 eV. We need to convert it into Joules using the relation: 1 eV = 1.6 x 10^{-19} J So, 2.3 eV = 2.3 * (1.6 x 10^{-19}) J = 3.68 x 10^{-19} J
04

Apply Planck's equation and find the threshold wavelength for sodium

Using the equation Φ = h*c/λ for sodium, we have: 3.68 x 10^{-19} J = (6.626 x 10^{-34} Js) * (3 x 10^8 m/s) / λ_sodium Solving for λ_sodium, we get: λ_sodium = 5.460 x 10^{-7} meters (which confirms our initial value)
05

Calculate the work function of tungsten in Joules

The work function of tungsten is given as 4.5 eV. We need to convert it into Joules using the relation: 4.5 eV = 4.5 * (1.6 x 10^{-19}) J = 7.2 x 10^{-19} J
06

Apply Planck's equation and find the threshold wavelength for tungsten

Using the equation Φ = h*c/λ for tungsten, we have: 7.2 x 10^{-19} J = (6.626 x 10^{-34} Js) * (3 x 10^8 m/s) / λ_tungsten Solving for λ_tungsten, we get: λ_tungsten = 3.102 x 10^{-7} meters
07

Convert the threshold wavelength of tungsten to Å

To convert the threshold wavelength of tungsten back to Å, we have: 1 meter = 10^10 Å So, 3.102 x 10^{-7} meters = 3.102 x 10^{-7} * 10^10 Å = 3102 Å Therefore, the threshold wavelength for tungsten is approximately 3102 Å, which is close to option (C) 2791 Å. This is not an exact match, so keep in mind that there may be some rounding error in the provided options.

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