Light of wavelength λ strikes a photoelectric surface and electrons are ejected with energy E . If E is to be increased to twice the original value, the wavelength changes to λ 1

Question

Light of wavelength λ strikes a photoelectric surface and electrons are ejected with energy E . If E is to be increased to twice the original value, the wavelength changes to λ 1 ​

TestHub · Accepted Answer

According to Einstein's photoelectric equation, the maximum kinetic energy $E$ of the ejected electrons is given by:

$E = \frac{h c}{λ} - W$

where $W$ is the work function of the material.

When the energy is increased to $2 E$ , the new wavelength is $λ_{1}$ :

$2 E = \frac{h c}{λ _{1}} - W$

From the first equation, we can write $E = \frac{h c}{λ} - W$ . Substituting this into the second equation gives:

$2 (\frac{h c}{λ} - W) = \frac{h c}{λ _{1}} - W$

$\frac{2 h c}{λ} - 2 W = \frac{h c}{λ _{1}} - W$

$\frac{h c}{λ _{1}} = \frac{2 h c}{λ} - W$

Dividing both sides by $h c$ yields:

$\frac{1}{λ _{1}} = \frac{2}{λ} - \frac{W}{h c}$

Since the work function $W$ is a positive quantity ( $W > 0$ ), we have:

$\frac{1}{λ _{1}} < \frac{2}{λ}$

Taking the reciprocal of both sides (since both sides are positive) gives:

$λ_{1} > \frac{λ}{2}$

Answer: $λ_{1} > λ /2$

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