Question

What is the wavelength of a photon that has the same momentum as an electron moving with a speed of 1500 m>s

Answer 1

Approximately
`4.85 * 10^(-7)\; {\rm m}` (approximately
`485\; {\rm nm}`.)

Step-by-step explanation:

The wavelength of this photon can be found in the following steps:

• Find the momentum of the given electron.
• Apply the relationship between the wavelength and momentum of a photon to find the required wavelength.

Since the electron in this question is travelling at a relatively low velocity of
`v = 1500\; {\rm m\cdot s^(-1)}`, the momentum
`p` of this electron can be approximated as:

`m_(e) \, v`,

Where
`m_(e) \approx 9.109 * 10^(-31)\; {\rm kg}` is the electron rest mass (mass of an electron that isn't moving.)

Substitute in
`v = 1500\; {\rm m\cdot s^(-1)}` to obtain:

`\begin{aligned}p &\approx m_{\text{e}}\, v \\ &\approx (9.109 * 10^(-31)\; {\rm kg})\, (1500\; {\rm m\cdot s^(-1)}) \\ &\approx 1.366 * 10^(-27)\; {\rm kg\cdot m\cdot s^(-1)}\end{aligned}`.

The momentum
`p` of a photon is inversely proportional to its wavelength
`\lambda`:

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

Where
`h \approx 6.626 * 10^(-34)\; {\rm kg \cdot m^(2)\cdot s^(-1)}` is the Planck constant.

Rearrange this equation to find wavelength
`\lambda` in terms of momentum
`p`:

`\begin{aligned}\lambda &= (h)/(p) \\ &\approx \frac{6.626 * 10^(-34)\; {\rm kg\cdot m^(2)\cdot s^(-1)}}{1.366 * 10^(-27)\; {\rm kg\cdot m\cdot s^(-1)}} \\ &\approx 4.85 * 10^(-7)\; {\rm m}\end{aligned}`.