Question

The French physicist Louis de Broglie in 1924 postulated that matter, like radiation, should exhibit a dual behaviour. He proposed the following relationship between the wavelength lambda of a material particle, its linear momentum p and planck constant h.
$\lambda =\frac{h}{p}=\frac{h}{mv}$
The de Broglie relation implies that the wavelength of a particle should decrease as its velocity increases. It also implies that for a given velocity heavier particles should have shorter wavelength than lighter particles. The waves associated with particles in motion are called matter waves or de Broglie waves. These waves differ from the electromagnetic waves as they
$\left(i\right)$ have lower velocities
$\left(ii\right)$ have no electrical and magnetic fields and
$\left(iii\right)$ are not emitted by the particle under consideration.
The experimental confirmation of the de Broglie’s relation was obtained when Davisson and Germer, in 1927, observed that a beam of electrons is diffracted by a nickel crystal. As diffraction is a characteristic property of waves, hence the beam of electron behaves as a wave, as proposed by de Broglie.

Using Bohr’s theory, the transition, so that the electron's de-Broglie wavelength becomes 3 times of its original value in $H{e}^{+}$ ion will be?

• A. $2⟶6$
• B. $2⟶4$
• C. $1⟶4$
• D. $1⟶6$

Answer 1

The correct option is A $2⟶6$

According to de Broglie relation:
$\lambda =\frac{h}{mv}$

$\lambda \propto \frac{1}{v}$
So, when the value of wavelength is increased to three times the velocity decreases to $1/3^{}$ times.
According to Bohr's model:
$v\propto \frac{z}{n}$
where $z=$ atomic number
$n=$ orbit number
If velocity is reduced to 1/3 times its original value $n$ has to become 3 times.
So, from the above option only option $\left(a\right)$ has $n$ value three times the original value.