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The critical wavelength for producing the photoelectric effect in tungsten metal is 2600Å. What wavelength would be necessary to produce photoelectrons from tungsten having twice the kinetic energy of those produced at 2200 Å?

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1800Å

1907Å

1926Å

2015Å

answer is B.

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Detailed Solution

Critical wavelength corresponds to work function.

$∴$ Work function $h v_{0} = \frac{h c}{λ_{0}}$

Also, KE at $λ_{1} = 2 (K E) a t 2200 Å$

$(\frac{h c}{λ_{1}} - \frac{h c}{λ_{0}}) = 2 (\frac{h c}{λ_{2}} - \frac{h c}{λ_{0}})$

$(\frac{1}{λ_{1}} - \frac{1}{6000}) = 2 (\frac{1}{2200} - \frac{1}{2600})$

$\frac{1}{λ_{1}} = \frac{1}{1100} - \frac{1}{2600}$

$\frac{1}{λ_{1}} = \frac{1500}{1100 \times 2600}$

$λ_{1} = \frac{2600 \times 11}{15} = 1907 Å$

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