Question

question 5
what is the maximum wavelength of light that can emit electrons from the surface of zn metal? the work function for zn is 6.91 × 10^{-19} j.
694 nm
121 nm
452 nm
288 nm

Explanation:

Step1: Recall the formula for the energy of a photon

The energy of a photon is given by \(E = h
u=\frac{hc}{\lambda}\), where \(h = 6.626\times10^{-34}\space J\cdot s\) (Planck's constant), \(c = 3\times10^{8}\space m/s\) (speed of light), and \(\lambda\) is the wavelength.
When the photon just has enough energy to emit an electron (i.e., the kinetic energy of the electron is zero), the energy of the photon \(E\) is equal to the work function \(\phi\). So, \(\phi=\frac{hc}{\lambda}\).

Step2: Solve for \(\lambda\)

Rearrange the formula \(\lambda=\frac{hc}{\phi}\).
Substitute \(h = 6.626\times10^{-34}\space J\cdot s\), \(c = 3\times10^{8}\space m/s\), and \(\phi=6.91\times 10^{-19}\space J\) into the formula:

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Convert meters to nanometers: \(1\space m=10^{9}\space nm\), so \(\lambda = 288\space nm\)

Answer:

288 nm