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Q12PE

Expert-verifiedFound in: Page 1063

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**What is the wavelength of EM radiation that ejects \(2.00\,{\rm{eV}}\) electrons from calcium metal, given that the binding energy is \(2.71\,{\rm{eV}}\)? What type of EM radiation is this?**

Wavelength of the radiation is \(263 \times {10^{ - 9}}\,{\rm{m}}\)

Type of EM radiation is Ultra violet.

**The kinetic energy of the electron is given by**

\(K{E_e} = hf - BE\) ...(1)

**Here **\(K{E_e}\)** is the kinetic energy, **\(h\)** is the plank constant, **\(f\)**is the frequency of the EM radiation and **\(BE\)** is the binding energy.**

**Now we know that the wavelength of EM radiation is given by**

\(\lambda = \frac{c}{f}\) ...(2)

**Where **\(c\)** is the speed of light.**

**So equation becomes,**

\(K{E_e} = \frac{{hc}}{\lambda } - BE\) ...(3)

The total energy of the photon is

\(\begin{aligned}{}E &= (2.00\,{\rm{eV}}) + (2.71\,{\rm{eV}})\\ &= 4.71\,{\rm{eV}}\end{aligned}\)

We also know that: Planks constant \(h = 4.13 \times {10^{ - 15}}\,{\rm{eV}}{\rm{.s}}\)

Speed of light \(c = 3 \times {10^8}\,{\rm{m/s}}\)

The wavelength of the radiation is expressed as,

\(\lambda = \frac{{hc}}{E}\)

Substitute all the value in the above equation

\(\begin{aligned}{}\lambda &= \frac{{\left( {4.13 \times {{10}^{ - 15}}\,{\rm{eV}}{\rm{.s}}} \right)\left( {3 \times {{10}^8}\,{\rm{m/s}}} \right)}}{{(4.71\,{\rm{eV}})}}\\ &= 263 \times {10^{ - 9}}\,{\rm{m}}\end{aligned}\)

Therefore, the wavelength of the radiation is \(263 \times {10^{ - 9}}\,{\rm{m}}\)

Hence, the wavelength of the EM radiation is \(263\,{\rm{nm,}}\)the radiation type is ultra violet.

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