A neutron moving with an energy of 20-4 ev collides with a stationary hydrogen atom Mp - m n - which is in ground state- Then due to impact the hydrogen atomA- gets excited to first excited stateB- gets excited to second excited stateC- neither gets excited or nor gets ionizedD- gets ionized

The correct option is A gets excited to first excited state1/2mU^2 =1/2(2m)V^2 +Δ E mU =2mV; K_i =2 Δ E Δ E =10.2 eV ∴ hydrogen atom gets exited to = 2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Multi-Electron Configurations and Orbital Diagrams

A neutron mov...

Question

A neutron moving with an energy of 20.4 ev collides with a stationary hydrogen atom $(M_{P} = m_{n})$ , which is in ground state. Then due to impact the hydrogen atom

gets ionized

No worries! We‘ve got your back. Try BYJU‘S free classes today!

gets excited to first excited state

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

gets excited to second excited state

No worries! We‘ve got your back. Try BYJU‘S free classes today!

neither gets excited or nor gets ionized

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B gets excited to first excited state
$\frac{1}{2} m U^{2} = \frac{1}{2} (2 m) V^{2} + Δ E m U = 2 m V; K_{i} = 2 Δ E Δ E = 10.2 e V$
$∴$ hydrogen atom gets exited to = 2

Suggest Corrections

Similar questions

Q. A beam of monochromatic light of wavelength λ ejects photoelectrons from a cesium surface (Φ = 1.9 eV). These photoelectrons are made to collide with hydrogen atoms in ground state. Find the maximum value of λ for which (a) hydrogen atoms may be ionized, (b) hydrogen atoms may get excited from the ground state to the first excited state and (c) the excited hydrogen atoms may emit visible light.

Q. A hydrogen sample is prepared in a particular excited state

A

. Photons of energy

2.55 e V

gets absorbed in the sample to take some electrons to a further excited state

B

. What are the quantum numbers of states

A

and

B

Q. The ionization energy of hydrogen atom is

13.6

eV. The energy required to excite the electron in a hydrogen atom from the ground state to the first excited state is:

Q. An electron collides with a fixed hydrogen atom in its groundstate. Hydrogen atom gets excited and the colliding electron losesall its kinetic energy. Consequently the hydrogen atom may emit aphoton corresponding to the largest wavelength of the Balmer series. The K.E. of colliding electron will be

Q. The first excited state of hydrogen atom is higher to its ground energy level by

10.2 e V

. The temperature necessary to excite hydrogen atom to first excited state will be

Join BYJU'S Learning Program

A neutron moving with an energy of 20.4 ev collides with a stationary hydrogen atom (MP=mn), which is in ground state. Then due to impact the hydrogen atom

The correct option is B gets excited to first excited state12mU2=12(2m)V2+ΔEmU=2mV;Ki=2ΔEΔE=10.2eV ∴ hydrogen atom gets exited to = 2

A neutron moving with an energy of 20.4 ev collides with a stationary hydrogen atom $(M_{P} = m_{n})$ , which is in ground state. Then due to impact the hydrogen atom

The correct option is B gets excited to first excited state
$\frac{1}{2} m U^{2} = \frac{1}{2} (2 m) V^{2} + Δ E m U = 2 m V; K_{i} = 2 Δ E Δ E = 10.2 e V$
$∴$ hydrogen atom gets exited to = 2