Question

For crystal diffraction experiments, wavelengths on the order of 0.25 nm are often appropriate.
A) Find the energy in electron volts for a particle with this wavelength if the particle is a photon.
B) Find the energy in electron volts for a particle with this wavelength if the particle is an electron.
C) Find the energy in electron volts for a particle with this wavelength if the particle is an alpha particle (m=6.64×10−27kg).

Answer 1

Answer:

A) E = 4.96 x 10³ eV

B) E = 4.19 x 10⁴ eV

C) E = 3.73 x 10⁹ eV

Explanation:

A)

For photon energy is given as:

$E = hv$

$E = \frac{hc}{\lambda}$

where,

E = energy of photon = ?

h = 6.625 x 10⁻³⁴ J.s

λ = wavelength = 0.25 nm = 0.25 x 10⁻⁹ m

Therefore,

$E = \frac{(6.625 x 10^{-34} J.s)(3 x 10^8 m/s)}{0.25 x 10^{-9} m}$

$E = (7.95 x 10^{-16} J)(\frac{1 eV}{1.6 x 10^{-19} J})$

E = 4.96 x 10³ eV

B)

The energy of a particle at rest is given as:

$E = m_{0}c^2$

where,

E = Energy of electron = ?

m₀ = rest mass of electron = 9.1 x 10⁻³¹ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

$E = (9.1 x 10^{-31} kg)(3 x 10^8 m/s)^2$

$E = (8.19 x 10^{-14} J)(\frac{1 eV}{1.6 x 10^{-19} J})$

E = 4.19 x 10⁴ eV

C)

The energy of a particle at rest is given as:

$E = m_{0}c^2$

where,

E = Energy of alpha particle = ?

m₀ = rest mass of alpha particle = 6.64 x 10⁻²⁷ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

$E = (6.64 x 10^{-27} kg)(3 x 10^8 m/s)^2$

$E = (5.97 x 10^{-10} J)(\frac{1 eV}{1.6 x 10^{-19} J})$

E = 3.73 x 10⁹ eV

Answer 2

A) The energy in electron volts for a particle with this wavelength if the particle is a photon is; .E = 4969.5 eV or 4.9695 keV

B) The energy in electron volts for a particle with this wavelength if the particle is an electron is; E = 23.58 eV

C) E = 0.003306 eV

A) The formula for the energy here is;

E = hc/λ

where;

h is planck's constant = 6.626 × 10⁻³⁴ J.s

c is speed of light = 3 × 10⁸ m/s

λ is wavelength = 0.25 nm = 0.25 x 10⁻⁹ m

Thus;

E = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(0.25 x 10⁻⁹)

79.512 × 10⁻¹⁷ J

converting to eV gives;

E = (79.512 × 10⁻¹⁷)/(1.6 × 10⁻¹⁹)

E = 4969.5 eV or 4.9695 keV

B) Formula for the energy if the particle is an electron is;

E = h²/(2mλ²)

where m = 9.31 × 10⁻³¹ kg

E = (6.626 × 10⁻³⁴)²/(2 × 9.31 × 10⁻³¹ × (0.25 x 10⁻⁹)²)

E = 37.726 × 10⁻¹⁹ J

Converting to eV gives;

E = (37.726 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹)

E = 23.58 eV

C) Mass of alpha particle is; m = 6.64 × 10⁻²⁷ kg

E = h²/(2mλ²)

where m = 6.64 × 10⁻²⁷ kg

E = (6.626 × 10⁻³⁴)²/(2 × 6.64 × 10⁻²⁷ × (0.25 x 10⁻⁹)²)

E = 52.896 × 10⁻²³ J

Converting to eV gives;

E = (52.896 × 10⁻²³)/(1.6 × 10⁻¹⁹)

E = 0.003306 eV

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