Question

What wavelength photon would have the same energy as a 145-gram baseball moving 30.2 m/s ?

Answer 1

Approximately
`3.01 * 10^(-27)\; {\rm m}` (when measured in a vacuum.)

Step-by-step explanation:

Apply unit conversion and ensure that the mass of the baseball is in standard units (kilograms):

`m = 145\; {\rm g} = 0.145\; {\rm kg}`.

The kinetic energy of the baseball will be:

`\displaystyle E = (1)/(2)\, m\, v^(2)`,

Where
`v = 30.2\; {\rm m\cdot s^(-1)}` is the speed of the baseball.

`\begin{aligned}E &= (1)/(2)\, m\, v^(2) \\ &= (1)/(2)\, (0.145)\, (30.2)^(2)\; {\rm J} \\ &= 66.12290\; {\rm J}\end{aligned}`.

The energy of a photon of frequency
`f` is:

`E = h\, f`,

Where
`h \approx 6.62607 * 10^(-34)\; {\rm m^(2)\cdot kg \cdot s^(-1)}` is Planck's constant.

When measured in a vacuum where speed of light is
`c \approx 3.00 * 10^(8)\; {\rm m\cdot s^(-1)}`, the wavelength
`\lambda` of this photon will be:

`\displaystyle \lambda = (c)/(f)`.

`\displaystyle f = (c)/(\lambda)`.

Hence, the expression for the energy of this photon can be rewritten as:

`\displaystyle E = h\, f = (h\, c)/(\lambda)`.

Rearrange this equation to find
`\lambda`:

`\displaystyle \lambda &= (h\, c)/(E)`.

Assuming that the energy of this photon to be equal to the kinetic energy of that baseball,
`66.12290\; {\rm J}`:

`\begin{aligned}\lambda &= (h\, c)/(E) \\ &\approx ((6.62607* 10^(-34))\, (3.00 * 10^(8)))/((66.12290))\; {\rm m} \\ &\approx 3.01 * 10^(-27)\; {\rm m}\end{aligned}`.