Question

Find the de Broglie wavelength of photoelectrons
ejected with maximum kinetic energy, when light of
wavelength 100 nm is incident on a cesium surface.
(Work function of cesium = 3.4 eV)​

Answer 1

The de Broglie wavelength of photoelectrons ejected with maximum kinetic energy, when light of wavelength 100 nm is incident on a cesium surface is λ = 4.08 A⁰

• The de-broglie wavelength is represented as:

        λ = h/mv

• As per the question, wavelength = 100 nm

                                                               = 100 x10⁻⁹ m

• Now, we know that,

        $\frac{hc}{\lambda} = \phi + KE$            (Einstein's equation)

• Thus, upon putting the values, we get,

$\frac{1240}{100} eV = 3.4 eV + KE$

12.4 eV = 3.4 eV +KE

KE = 9eV

• Now, the de-broglie wavelength will be :

$\lambda = \frac{12.25}{\sqrt{9}}A^o$

λ = 4.08 A⁰

Answer 2

Answer:

De-Broglie wavelength of the photoelectrons of maximum kinetic energy is $\lambda = 4.08 A^o$

Explanation:

As we know that de broglie wavelength of the accelerated electron is given as

$\lambda = \frac{12.25 A^o}{\sqrtV}$

now we can use Einstein's equation

$\frac{hc}{\lambda} = \phi + KE$

so we have

$\frac{1240}{100} eV = 3.4 eV + KE$

$12.4 eV - 3.4 eV = KE$

$KE = 9 eV$

now from above formula we have

$\lambda = \frac{12.25}{\sqrt{9}}A^o$

$\lambda = 4.08 A^o$

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Topic : De Broglie Wavelength

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