The threshold frequency of silver is 9.71×1015 Hz. What is the work function of silver?

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The threshold frequency of silver is 9.71×1015 Hz. What is the work function of silver?

Planck’s constant is 6.63 × 10−34 J · s. Answer in units of eV.

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Answer 1 · 2018-05-16T22:55:54

$W = 40.2$ $W = 40.2$ $eV$ $"eV"$

Explanation:

Threshold frequency is related to the work function by the following equation:

$f_{0} = \frac{W}{h}$ $f_(0) = frac(W)(h)$ ; where $f_{0}$ $f_(0)$ is the threshold frequency, $W$ $W$ is the work function, and $h$ $h$ is the Planck constant.

We can rearrange this equation to solve for $W$ $W$ :

$\Rightarrow W = f_{0} h$ $Rightarrow W = f_(0) h$

Let's plug in the given values:

$\Rightarrow W = (9.71 \times 10^{15} \times 6.63 \times 10^{- 34})$ $Rightarrow W = (9.71 times 10^(15) times 6.63 times 10^(- 34))$ $J$ $"J"$

$\Rightarrow W = 6.43773 \times 10^{- 18}$ $Rightarrow W = 6.43773 times 10^(- 18)$ $J$ $"J"$

However, we need to answer in units of $eV$ $"eV"$ :

$\Rightarrow W = (\frac{6.43773 \times 10^{- 18}}{1.6 \times 10^{- 19}})$ $Rightarrow W = (frac(6.43773 times 10^(- 18))(1.6 times 10^(- 19)))$ $eV$ $"eV"$

$therefore W approx 40.2$ $"eV"$

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The threshold frequency of silver is 9.71×1015 Hz. What is the work function of silver?

Planck’s constant is 6.63 × 10−34 J · s. Answer in units of eV.

1 Answer

Explanation:

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