A monochromatic light source of frequency $f$ illuminates a metallic surface and ejects photoelectrons. The photo electrons having maximum energy are just able to ionize the hydrogen atoms in ground state. When the entire experiment is repeated with an incident radiation of frequency $\frac{5}{6} f$ , the photoelectrons so emitted are able to excite the hydrogen atom which then emits a radiation of wavelength $1215 \overset{\circ}{A}$ . Find the work function of the metal.

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Infinite

$6.8 e v$

$4 e v$

Zero

answer is A.

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Detailed Solution

Using photoelectricity equation,

$h f - ϕ_{0} = 13.6 eV \dots (i)$

$h (\frac{5}{6} f) - ϕ_{0} = \frac{12375}{1215} = 10.2 eV \dots (i i)$

Subtracting (ii) from (i),

$\frac{h f}{6} = 13.6 - 10.2 = 3.4 eV \Rightarrow h f = 6 \times 3.4 = 20.4 eV$

From equation (i), we have

$ϕ_{0} = h f - 13.6 = 20.4 - 13.6 = 6.8 eV$

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