The wavelength of an X-ray beam is 10 \AA. The mass of a fictitious particle having the same energy

Q: Why do we use $E=mc^2$ and not kinetic energy for the particle?

The problem asks for the mass of a 'fictitious particle' having the same energy. Unless a velocity is specified, we assume the energy referred to is its rest mass energy, which is the intrinsic energy associated with its mass.

Q: How do I convert Angstroms to meters?

One Angstrom ($\text{\AA}$) is equal to $10^{-10}$ meters. So, $10 \text{ \AA}$ is $10 \times 10^{-10} \text{ m}$, which simplifies to $10^{-9} \text{ m}$.

Q: Why didn't we substitute the numerical value of Planck's constant ($h$)?

Look closely at the format of the answer requested in the question: $\frac{x}{3}h$. The final answer is expected to be in terms of $h$, so substituting its numerical value would only complicate the calculation unnecessarily.

Q: What happens to the $c$ in the numerator of the photon energy equation?

When we equate $\frac{hc}{\lambda} = mc^2$, we divide both sides by $c^2$ to isolate $m$. This gives $m = \frac{hc}{\lambda c^2}$. The $c$ in the numerator cancels with one $c$ in the denominator, leaving $m = \frac{h}{c\lambda}$.

Q: Is this a real particle?

No, the question explicitly calls it a 'fictitious particle'. It's a theoretical exercise designed to test your understanding of equating wave energy (photon) with mass energy.

Question

The wavelength of an X-ray beam is $10 	ext{ \AA}$. The mass of a fictitious particle having the same energy as that of the X-ray photons is $\frac{x}{3}h 	ext{ kg}$. The value of $x$ is ............ .
($h = 	ext{Planck's constant}$)

Sigma Percentile · Accepted Answer

This problem is a beautiful demonstration of the deep connection between the wave nature of light and the particle nature of matter, bridged by the concept of energy.

### Analyzing the Setup

We are given an X-ray photon with a specific wavelength, $\lambda = 10 	ext{ \AA}$. Alongside it, we have a "fictitious particle" that possesses the exact same amount of energy as the photon. Our goal is to find the mass of this particle in terms of Planck's constant, $h$.

To begin, we need to express the energy of both entities. For the X-ray photon, its energy is governed by the Planck-Einstein relation:

$$E_{	ext{photon}} = \frac{hc}{\lambda}$$

For the fictitious particle, assuming it is at rest (as no velocity is mentioned), its energy is given by Einstein's famous mass-energy equivalence principle:

$$E_{	ext{particle}} = mc^2$$

### The Master Equation

Since the problem states that their energies are equal, we can set these two expressions equal to each other:

$$\frac{hc}{\lambda} = mc^2$$

We want to find the mass, $m$. Let's rearrange the equation to isolate $m$. Notice that one factor of the speed of light, $c$, cancels out from both sides:

$$m = \frac{h}{c\lambda}$$

### Final Calculation

Now, it's time to substitute the known values. It is absolutely critical to convert all units to the standard SI system to avoid errors. The wavelength is given in Angstroms, so we convert it to meters:

$$\lambda = 10 	ext{ \AA} = 10 	imes 10^{-10} 	ext{ m} = 10^{-9} 	ext{ m}$$

The speed of light in a vacuum is a standard constant:

$$c = 3 	imes 10^8 	ext{ m/s}$$

Substituting these into our isolated equation for mass:

$$m = \frac{h}{(3 	imes 10^8 	ext{ m/s}) 	imes (10^{-9} 	ext{ m})}$$

Let's simplify the denominator. Multiplying the powers of 10, we get $10^8 	imes 10^{-9} = 10^{-1}$:

$$m = \frac{h}{3 	imes 10^{-1}}$$

Bringing the $10^{-1}$ from the denominator up to the numerator changes its sign, making it $10^1$, or simply $10$:

$$m = \frac{10h}{3} 	ext{ kg}$$

The problem states that the mass of the particle is $\frac{x}{3}h 	ext{ kg}$. By comparing our derived expression with the given format, it is clear that:

**$$x = 10$$**

This elegant problem reminds us of the profound symmetry in physics, where mass and energy are simply two sides of the same coin.

Enter Numerical Value:

Visualized Solution

$E_{photon} = E_{particle}$

$Energy Formulas$

$\frac{h c}{λ} = m c^{2}$

$Substitute Values$

$Simplify Denominator$

$m = \frac{10}{3} h$

$Conclusion$

The Sigma Insight: Photon Theory of Light

Analyzing the Setup

The Master Equation

Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ephoton​=Eparticle​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy Formulas

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λhc​=mc2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitute Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplify Denominator

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m=310​h

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion