The emitter in a photoelectric tube has a threshold wavelength of 6000 Å. Determine the wavelength of the light incident on the tube if the stopping potential for this light is 2.5 V.

Given, threshold wavelength,

λ_{o} =

6000 Å
Stopping potential for the light, V_o = 2.5 V

Now, the formula gives us,

The work function is,

ϕ_{0} = {hv}_{th} = \frac{hc}{λth} = \frac{12.4 \times 10^{3} eV \times Å}{6000 Å} = 2.07 eV

The photoelectric equation then gives,

{eV}_{0} = hv - ϕ_{0}

\Rightarrow

{eV}_{0} = \frac{hc}{λ} - ϕ_{0}

\Rightarrow

2.5 eV = \frac{12.4 \times 10^{3} eV . Å}{λ} - 2.07 eV

i.e.,

λ = 2713 \overset{o}{A} .

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