Question

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.
452 nm
288 nm
121 nm
694 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 of light.

When the photon just has enough energy to emit an electron (i.e., the kinetic energy of the emitted electron \(K = 0\)), the energy of the photon \(E\) is equal to the work function \(\phi\) of the metal. So, \(E=\phi\).

We can rewrite the formula for \(\lambda\) as \(\lambda=\frac{hc}{\phi}\)

Step2: Substitute the values into the formula

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

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Convert meters to nanometers: Since \(1\space m = 10^{9}\space nm\), \(\lambda=2.88\times 10^{-7}\times10^{9}\space nm = 288\space nm\)

Answer:

288 nm