If the de-Broglie wavelength of the electron in nth Bohr orbit in a hydrogenic atom is equal to 1.5

Q: Why must the circumference be exactly $n\lambda$?

If the circumference is not an exact integer multiple of the wavelength, the wave will not perfectly overlap with itself after one revolution. Over multiple revolutions, the peaks and troughs will misalign, causing destructive interference and canceling the wave out. For the electron to exist stably, it must form a constructive standing wave.

Q: What exactly is $a_0$ in the radius formula?

$a_0$ is the Bohr radius, which is approximately $0.529 \text{ \AA}$. It represents the most probable distance between the nucleus and the electron in a ground-state hydrogen atom ($n=1, Z=1$).

Q: Could I have solved this using Bohr's angular momentum quantization instead?

Absolutely! Bohr's postulate states $mvr = \frac{nh}{2\pi}$. If you substitute the de-Broglie relation $\lambda = \frac{h}{mv}$ (which means $mv = \frac{h}{\lambda}$) into Bohr's postulate, you get $\left(\frac{h}{\lambda}\right)r = \frac{nh}{2\pi}$, which rearranges exactly to $2\pi r = n\lambda$.

Q: Is it mathematically safe to cancel $n$ from both sides of the equation?

Yes. In algebra, you cannot divide by zero. However, $n$ represents the principal quantum number (the orbit number), which takes integer values $1, 2, 3, \dots$. Since $n$ can never be zero, dividing both sides by $n$ is perfectly valid.

Q: What does the ratio $n/Z = 0.75$ actually tell us physically?

It gives us a constraint on which atoms and which orbits can have this specific wavelength. For example, since $n$ and $Z$ must be integers, $n/Z = 3/4$ means this condition is met when an electron is in the $3^{\text{rd}}$ orbit ($n=3$) of a Beryllium ion ($Z=4$, $\text{Be}^{3+}$).

Question

If the de-Broglie wavelength of the electron in $n^{	ext{th}}$ Bohr orbit in a hydrogenic atom is equal to $1.5 \pi a_0$ ($a_0$ is Bohr radius), then the value of $n/Z$ is

Options:

(A) 1.0

(B) 0.75

(C) 0.40

(D) 1.50

Sigma Percentile · Accepted Answer

### The Wave Nature of the Electron

Imagine an electron not as a tiny billiard ball orbiting the nucleus, but as a continuous, vibrating wave. This is the heart of Louis de-Broglie's revolutionary hypothesis. For an electron to exist in a stable Bohr orbit without its wave interfering destructively with itself, the wave must perfectly close on itself.

Geometrically, this means the circumference of the circular orbit must be an exact integer multiple of the electron's de-Broglie wavelength. We can write this profound physical condition as a simple mathematical equation:

$$2\pi r = n\lambda$$

Here, $r$ is the radius of the orbit, $n$ is the principal quantum number (which tells us the number of full waves), and $\lambda$ is the de-Broglie wavelength.

### Bringing in Bohr's Radius

To solve our problem, we need to express the radius $r$ in terms of known constants. From Bohr's model of hydrogenic (one-electron) atoms, the radius of the $n^{	ext{th}}$ orbit is given by:

$$r = \frac{a_0 n^2}{Z}$$

In this formula, $a_0$ is the fundamental Bohr radius (the radius of the first orbit in a hydrogen atom), and $Z$ is the atomic number of the nucleus.

Let's substitute this expression for $r$ back into our standing wave equation. This merges the wave nature of the electron with the structural geometry of the atom:

$$2\pi \left(\frac{a_0 n^2}{Z}ight) = n\lambda$$

### The Final Calculation

The problem provides us with a very specific value for the de-Broglie wavelength: $\lambda = 1.5 \pi a_0$. Let's plug this directly into our merged equation:

$$2\pi a_0 \frac{n^2}{Z} = n(1.5 \pi a_0)$$

Now, we get to enjoy the elegance of algebra. Notice how many terms appear on both sides of the equation. We can safely divide both sides by $n$ (since the orbit number $n$ is always a positive integer, never zero). We can also cancel out the $\pi$ and the $a_0$ terms.

After clearing the clutter, we are left with:

$$2 \frac{n}{Z} = 1.5$$

To find the value of the ratio $\frac{n}{Z}$, we simply divide by 2:

$$\frac{n}{Z} = \frac{1.5}{2} = 0.75$$

And there we have it! By combining the geometry of standing waves with Bohr's radius formula, we've elegantly arrived at the final answer.

Animated Solution for Chemistry - Atomic Structure: If the de-Broglie wavelength of the electron in $n^{th}$ Bohr orbit in a hydrogenic atom is equal to $1.5 π a_{0}$ ( $a_{0}$ is Bohr radius), then the value of $n / Z$ is

Select Answer:

Visualized Solution

$2 π r = nλ$

$r = \frac{a _{0} n ^{2}}{Z}$

$2 π (\frac{a _{0} n ^{2}}{Z}) = nλ$

$λ = 1.5 π a_{0}$

$\frac{n}{Z} = 0.75$

$Alternative Approach$

The Sigma Insight: Wave Particle Duality

The Wave Nature of the Electron

Bringing in Bohr's Radius

The Final Calculation

Similar Questions

The de-Broglie wavelength ( $λ$ ) associated with a photoelectron varies with the frequency ( $ν$ ) of the incident radiation as, [ $ν_{0}$ is threshold frequency]

A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference $V$ esu. If $e$ and $m$ are charge and mass of an electron, respectively, then the value of $h / λ$ (where, $λ$ is wavelength associated with electron wave) is given by

A 50 watt bulb emits monochromatic red light of wavelength of $795 nm$ . The number of photons emitted per second by the bulb is $x \times 1 0^{20}$ . The value of $x$ is ......... . [Given, $h = 6.63 \times 1 0^{- 34} Js$ and $c = 3.0 \times 1 0^{8} ms^{- 1}$ ]

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2πr=nλ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=Za0​n2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2π(Za0​n2​)=nλ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ=1.5πa0​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Zn​=0.75

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