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Light of wavelength lambda strikes a photosensitive surface and electrons are ejected with kinetic energy E. To double the kinetic energy to 2E, the wavelength must be changed to lambda' where:

1. lambda' = lambda/2
2. lambda' = 2*lambda
3. lambda/2 < lambda' < lambda
4. lambda' > lambda

Correct answer: lambda/2 < lambda' < lambda

Solution

From photoelectric effect: E = hc/lambda - phi...(1). 2E = hc/lambda' - phi...(2). Subtracting (1) from (2): E = hc/lambda' - hc/lambda => hc/lambda' = hc/lambda + E = hc/lambda + (hc/lambda - phi) = 2hc/lambda - phi. So 1/lambda' = (2/lambda) - phi/(hc). Since phi > 0, 1/lambda' < 2/lambda, meaning lambda' > lambda/2. Also, since E > 0 means hc/lambda > phi > 0, so hc/lambda' = hc/lambda + E > hc/lambda, meaning lambda' < lambda. Therefore lambda/2 < lambda' < lambda.

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