2
$\begingroup$

If the mass of an electron becomes 10 times its actual mass, which of the following statements is correct regarding Bohr's model:

  1. Velocity of electron increases by 10 times.
  2. Orbit radius decreases by 10 times.
  3. Energy of electron increases by 10 times.
  4. Wavelength of the electron remains the same.

(An objective question)

I first tried the rest mass formula, but it is not related to the options. So I tried the energy and radius formulae in Bohr's theory and they give option 4. But the wave nature of electrons is not discussed in Bohr's theory, is it?

The formulae:

Radius $r= n²h²/4π²mze²$
Total energy $E_n= -ze²/2r= -2ze²π²mze²/n²h²$
Velocity, from $MVR=nh/2π$ is $v=2πze²/nh$

$\endgroup$
3
  • 1
    $\begingroup$ But the wave nature of electrons is not discussed in Bohr's theory, is it? Well, Bohr theory considers the length of an electron orbit is an integer multiple of the electron de Broglie wavelength. $\endgroup$
    – Poutnik
    Jun 4 at 7:27
  • 1
    $\begingroup$ @Poutnik I think the temporal order is reversed. Then the way the exercise should be seen or used to learn depends on the stage at which it is proposed. $\endgroup$
    – Alchimista
    Jun 4 at 8:37
  • 1
    $\begingroup$ For an actual example, look up muonic hydrogen. $\endgroup$ Jun 6 at 22:26
3
$\begingroup$

Look for those quantities which may vary when mass of electron may vary i.e those quantities which contain term $m_e$ ( where $m_e$ is mass of electron) in their expression.

Velocity in $n^{th}$ orbit= $\dfrac{2kZe^2}{nh}$ (independent of mass)

Radius of $n^{th}$ orbit = $\dfrac{n^2h^2}{4k\pi^2Ze^2m_e}$ ( varies inversely with mass)

Energy of electron $n^{th}$ orbit= $\dfrac{-kZe^2}{2r_n}$ (varies proportionally with mass)

Wavelength of electron = $\dfrac{2\pi r_n}{n}$ (varies inversely with mass)

Thus, the velocity remains unchanged, radius decreases by 10 times, energy increases by 10 times and wavelength decreases by 10 times.

Therefore, the final answer would be (2) and (3).

$\endgroup$
0
2
$\begingroup$

If you're answering this in an exam setting, you likely don't remember the formulae for velocity and radius of electron in $n$th orbit (you should, however, learn those for energy and wavelength). The answer given by kanchan is correct but relies on memorization of unnecessary formula, so let me propose a logical way of thinking about this.

In Bohr's model, electrons travel in circular orbits around the nucleus. As centripetal force is always perpendicular to the electron's velocity, its angular momentum doesn't change. That is to say,

$$mvr=\text{constant} \tag{1}$$

This means that if mass increases by $10$ times and velocity stays constant (I'll prove that it does in a while), the radius of the orbit should decrease by $10$ times.


The centripetal force acting on the electron is almost entirely electrostatic (you can calculate the gravitational force between the hydrogen nucleus and its electron to convince yourself of this). So,

$$\frac{mv^2}{r}=\frac{1}{4\pi\epsilon_o}\cdot \frac{Ze^2}{r^2}$$

Multiplying both sides by $r^2$

$$mv^2r=\frac{1}{4\pi\epsilon_o}\cdot Ze^2 = \text{constant} \tag{2}$$

If you divide $(2)$ by $(1)$ you get

$$v=\text{constant}$$

This means that the velocity of an electron won't change with mass.


The formulae for energy and wavelength of electron (which, as I said, you should remember) are

$$E=-\frac{1}{4\pi\epsilon_o}\cdot \frac{Ze^2}{2r}$$ $$\lambda=\frac{2\pi r}{n}$$

You can see that energy is inversely and wavelength directly proportional to radius. But you also know that radius varies inversely as mass, so you should reverse the proportionalities of radius with energy and wavelength to get the corresponding proportionalities for mass with those quantities.


To answer your question, when mass of the electron dectuples, radius becomes smaller by a factor of $10$, velocity stays the same, energy increases by a factor of $10$ and wavelength decreases by a factor of $10$.

The correct options therefore are (2) and (3)

$\endgroup$
3
  • $\begingroup$ we should never call a formula unnecessary, formulas are meant to be remembered. @RayBradbury the questioner may not have enough time during his exam to think about such logics in exam , If one knows concept and has understood derivation, there's no harm in remembering formula if one can. The questioner himself has mentioned formulas and I think he has no harm in remembering them $\endgroup$
    – Lllt
    Jun 5 at 3:21
  • 2
    $\begingroup$ @kanchan I can only remember a handful of formulae for an exam. I want those that I remember to be short and of wide applicability. Indeed, I never memorised the formulae for velocity and radius and seemed to do okay. In my experience, it is easier to remember a short proof than arbitrary trivia ("radius shrinks as mass increases, velocity doesn't change"). But you may be right, and OP is free to discard my answer for yours. $\endgroup$ Jun 5 at 4:00
  • $\begingroup$ I have mentioned the formulae but I too have the inability to remember formulae if they're being very specific. Also, the above ones are quite simple to obtain using Coulomb's Law, not unlike @Ray's method here. $\endgroup$ Jun 5 at 9:32

Not the answer you're looking for? Browse other questions tagged or ask your own question.