If $λ_{1}$ and $λ_{2}$ denote the wavelength of de Broglie waves for electrons in Bohr’s first and second orbits in the hydrogen atom, then $\frac{λ_{1}}{λ_{2}}$ will be:

Q. The wavelength of photons emitted by electron transition between two similar levels in

H

and

H e^{+}

are

λ_{1}

and

λ_{2}

resepectively. Then the relation between

λ_{1} and λ_{2}

is:

Q. Two particles moving at right angles have their de-Broglie wavelengths

λ_{1}

and

λ_{2}

. If their collision is perfectly inelastic, what is the final wavelength

λ

, in terms of

λ_{1}

and

λ_{2} ?

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If λ1 and λ2 denote the de-Broglie wavelength of two particles with same masses but charges in the ratio of 1 : 2 after they are accelerated from rest through the same potential difference, then:

The correct option is B λ1>λ2λ=h√2mKE and KE=qV where q is the charge on the particle.Since the particles have same mass, λ1λ2=√KE2√KE1.KE1<KE2 and hence λ1>λ2.

If $λ_{1}$ and $λ_{2}$ denote the de-Broglie wavelength of two particles with same masses but charges in the ratio of 1 : 2 after they are accelerated from rest through the same potential difference, then:

The correct option is B $λ_{1} > λ_{2}$
$λ = \frac{h}{\sqrt{2 m K E}}$ and $K E = q V$ where $q$ is the charge on the particle.
Since the particles have same mass, $\frac{λ_{1}}{λ_{2}} = \frac{\sqrt{K E_{2}}}{\sqrt{K E_{1}}}$ .
$K E_{1} < K E_{2}$ and hence $λ_{1} > λ_{2}$ .