Question

Light of wavelength 310 nm strikes a metal whose work function is 2.3 ev. (a) What is the maximum kinetic energy of the ejected electrons? (b) What is the de Broglie wavelength for the electrons that are produced?

Answer 1

1. 71 eV

0.939 nm

Step-by-step explanation:

Data:

A. The wavelength of light =
`3 * 10^(-9) m`

work function = 2.3 eV

mass of an electron =
`9.1 * 10-^(31) kg`

the maximum kinetic energy is given by the following equation:

`E_(k) = hv - m_(o) \\ h = 6.626 * 10^(-34)`

therefore, Ek =
`(1987.8)/(496) - 2.3\\ = 1.71 eV`

B. the wavelength is given by:

p =
`\sqrt{2*9.1*10^(-31)* 1.71*1.6*10^(-19) }\\= 7.06+ * 10^(-25) ms`

using the de brogile wavelength equation:

`\lambda = (h)/(p)\\ = (6.626*10^(-34) )/(7.06*10^(-25) ) \\ = 0.939 m`

Answer 2

a.1.71 eV

b.0.939 nm

Step-by-step explanation:

We are given that

Wavelength of light =310 nm=
`310* 10^(-9)` m

Work function=2.3 eV

Mass of electron=
`9.1* 10^(-31)`Kg

`\\u=(c)/(\lambda)`

a.We have to find the maximum kinetic energy of ejected electron

`K.E=h\\u-w_0`

`h={6.626* 10^(-34)`

`K.E=(6.626* 10^(-34)* 3*10^8)/(310*10^(-9)*1.6* 10^(-19))-2.3`

`K.E=(1987.8)/(496)-2.3`

K.E=1.71 eV

Hence, the maximum kinetic energy of ejected electron=1.71 eV

b.Kinetic energy =
`(p^2)/(2m)`

p=
`\sqrt{2* 9.1* 10^(-31)* 1.71*1.6*10^(-19)}`

p=
`7.06* 10^(-25)`m-s

We know that de brogile wavelength

`\lambda =(h)/(p)`

`\lambda =(6.626*10^(-34))/(7.06* 10^(-25))`

`\lambda=0.939* 10^(-9)`

`\lambda=0.939 nm`

Hence, the de-brogile wavelength of ejected electron=0.939 nm.