If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of $1.5 \times 10^{7}\ m\ s^{-1}$, calculate the energy with which it is bound to the nucleus.

If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out wit

Exercises · 2.54

1.5 \times 10^{7}\ m\ s^{-1}

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Concept understanding — Energy Level Quantization

Energy Level Quantization: From Intuition to Precision

Imagine you're climbing a smooth ramp. You can stop at any height — 1 metre, 1.5 metres, 2.1 metres — anywhere you like. That's how we intuitively think about energy in everyday life: continuous, like a slide.

Now imagine a staircase. You can stand on step 1, step 2, or step 3 — but you cannot stand halfway between step 2 and step 3. There's no such place. The steps are discrete, not continuous.

Energy level quantization is the idea that in the microscopic world of atoms and molecules, energy behaves like a staircase, not a ramp. Electrons in an atom cannot have just any energy — they can only occupy specific, allowed energy levels. Everything else is forbidden.

Why does this happen? The core intuition

In the classical world, an electron orbiting a nucleus would continuously radiate energy, spiral inward, and crash — atoms would be unstable. But atoms are stable. Nature solved this problem by imposing a rule: the electron's angular momentum (and therefore its energy) can only take certain discrete values.

Think of a guitar string. It can only vibrate at specific frequencies — its fundamental and harmonics. You can't pluck it to produce a frequency halfway between two harmonics. The string's vibration is quantized by its boundaries. Similarly, an electron bound to a nucleus is confined in space, and that confinement forces its energy to be quantized.

Note

Quantization is not a mysterious extra rule — it emerges naturally whenever a wave (like an electron's matter wave) is confined. Confinement creates standing waves, and standing waves only exist at specific frequencies.

The precise statement

For a bound system (like an electron in an atom), the total energy $E$ of the system can only take certain discrete values:

E = E_1, E_2, E_3, \dots

where each $E_n$ is a specific, fixed number. The integer $n$ (1, 2, 3, …) is called the principal quantum number. The lowest energy level ( $n=1$ ) is the ground state; higher levels ( $n>1$ ) are excited states.

For the hydrogen atom, the allowed energies are given by:

E_n = -\frac{13.6 \text{ eV}}{n^2}

So:

$n=1$ : $E_1 = -13.6$ eV (ground state)
$n=2$ : $E_2 = -3.4$ eV
$n=3$ : $E_3 = -1.51$ eV
and so on, approaching 0 eV as $n \to \infty$ (the ionization limit)

Important

The negative sign means the electron is bound to the nucleus. Zero energy corresponds to the electron being free (ionized). The more negative the energy, the more tightly bound the electron.

How do we know this is real?

The most direct evidence comes from atomic spectra. When an electron jumps from a higher energy level to a lower one, it emits a photon of light with energy exactly equal to the difference:

\Delta E = E_{\text{higher}} - E_{\text{lower}} = hf

where $h$ is Planck's constant and $f$ is the frequency of the emitted light. …

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When a photon ejects an inner electron, its energy splits between overcoming the binding energy and giving the electron kinetic energy. For a 150 pm photon ejecting an electron at $1.5 \times 10^7$ m/s, the binding energy is $1.204 \times 10^{-15}$ J or 7.52 keV.

Why this approach works

When a high-energy photon collides with an atom, it can knock out an inner electron in a process similar to the photoelectric effect. The photon's energy doesn't vanish—it's conserved. Part of it goes into breaking the electron free from the nucleus (the binding energy we're hunting for), and whatever remains becomes the kinetic energy of the ejected electron.

This is pure energy accounting:

E_{\text{photon}} = E_{\text{binding}} + KE_{\text{electron}}

Rearranging gives us the binding energy directly once we know the photon's energy and the electron's kinetic energy.

Step-by-step solution

1. Calculate the photon's energy

The energy of a photon is tied to its wavelength through Planck's relation:

E_{\text{photon}} = \frac{hc}{\lambda}

where $h = 6.626 \times 10^{-34}$ J·s is Planck's constant, $c = 3 \times 10^8$ m/s is the speed of light, and $\lambda = 150$ pm $= 150 \times 10^{-12}$ m.

Substituting:

E_{\text{photon}} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{150 \times 10^{-12}}

E_{\text{photon}} = \frac{1.9878 \times 10^{-25}}{1.5 \times 10^{-10}} = 1.3252 \times 10^{-15} \text{ J}

2. Calculate the kinetic energy of the ejected electron

The electron emerges with velocity $v = 1.5 \times 10^7$ m/s. Its kinetic energy is:

KE = \frac{1}{2}mv^2

where $m = 9.109 \times 10^{-31}$ kg is the electron mass.

KE = \frac{1}{2}(9.109 \times 10^{-31})(1.5 \times 10^7)^2

KE = \frac{1}{2}(9.109 \times 10^{-31})(2.25 \times 10^{14})

KE = 1.025 \times 10^{-16} \text{ J}

3. Apply energy conservation to find binding energy

The binding energy is what's left after subtracting the kinetic energy from the photon's energy:

$E_{\text{binding}} = E_{\text{photon}} - KE$ …

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Q.If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5×107 m s−11.5 \times 10^{7}\ m\ s^{-1}1.5×107 m s−1, calculate the energy with which it is bound to the nucleus.

Why this approach works

Step-by-step solution

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Q.If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of $1.5 \times 10^{7}\ m\ s^{-1}$ , calculate the energy with which it is bound to the nucleus.

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