A particle P collides inelastically with a singly ionized helium atom which is at rest. Initially helium atom $(H e^{+})$ is in ground state. After collision particle ‘P’ is found to be scattered at an angle $90 °$ with respect to it’s initial direction of motion, and before emitting any radiation, $H e^{+}$ and particle P have same kinetic energy. The $H e^{+}$ atom after collision was found to be in such an excited state that the maximum possible number of different wavelengths representing the successive transitions in between excited level to ground level is 6. Mark the CORRECT option(s).

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

The particle P may be a sufficiently energetic electron

The particle P may be a sufficiently energetic neutron

Out of six wavelengths only one belongs to visible region spectrum

No emitted radiations belongs to visible region spectrum

answer is A, B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

The initial and final situations are marked using Conservation Of Linear Momentum.

In x direction, $m u + 0 = 0 + M v_{x}$

$\Rightarrow v_{x} = \frac{m u}{M}$

In y direction, 0+ $0 + 0 = m v - M v_{y}$

$\Rightarrow v_{y} = \frac{m v}{M}$
We are told, $K E_{m} = K E_{M}$ in final state.
$\begin{array}{l} \Rightarrow \frac{1}{2} m v^{2} = \frac{1}{2} M V^{2} \\ \Rightarrow \frac{1}{2} m v^{2} = \frac{1}{2} M \frac{m^{2}}{M^{2}} (u^{2} + v^{2}) \\ (∵ f r o m f i g, V = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{{(\frac{m u}{M})}^{2} + {(\frac{m v}{M})}^{2}}) \\ \Rightarrow v^{2} = \frac{m}{M} (u^{2} + v^{2}) \end{array}$
$\Rightarrow v^{2} = \frac{m u^{2}}{M - m}$ , which is possible for $M > m$
Hence particle P can be electron or neutron
From n = 7 to n = 1, 6 maximum possible wavelengths are there, of which n = 4 to n = 3 only belongs to visible spectrum.