Question

For crystal diffraction experiments, wavelengths on the order of 0.25 nm are often appropriate.

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