Question

What are (a) the energy of a photon corresponding to wavelength 6.0 nm, (b) the kinetic energy of an electron with de Broglie wavelength 6.0 nm, (c) the energy of a photon corresponding to wavelength 6.0 fm, and (d) the kinetic energy of an electron with de Broglie wavelength 6.0 fm?

Answer 1

Step-by-step explanation:

Given that,

Wavelength = 6.0 nm

de Broglie wavelength = 6.0 nm

(a). We need to calculate the energy of photon

Using formula of energy

`E = (hc)/(\lambda)`

`E=(6.63*10^(-34)*3*10^(8))/(6.0*10^(-9))`

`E=3.315*10^(-17)\ J`

(b). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

`\lambda=(h)/(√(2mE))`

`E=(h^2)/(2m\lambda^2)`

Put the value into the formula

`E=((6.63*10^(-34))^2)/(2*9.1*10^(-31)*(6.0*10^(-9))^2)`

`E=6.709*10^(-21)\ J`

(c). We need to calculate the energy of photon

Using formula of energy

`E = (hc)/(\lambda)`

`E=(6.63*10^(-34)*3*10^(8))/(6.0*10^(-15))`

`E=3.315*10^(-11)\ J`

(d). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

`\lambda=(h)/(√(2mE))`

`E=(h^2)/(2m\lambda^2)`

Put the value into the formula

`E=((6.63*10^(-34))^2)/(2*9.1*10^(-31)*(6.0*10^(-15))^2)`

`E=6.709*10^(-9)\ J`

Hence, This is the required solution.