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Y_Kistochka [10]
3 years ago
8

A mass m neutron has elastic collision with a mass m'

Physics
1 answer:
hoa [83]3 years ago
5 0

Answer:

The neutron loses all of its kinetic energy to nucleus.

Explanation:

Given:

Mass of neutron is 'm' and mass of nucleus is 'm'.

The type of collision is elastic collision.

In elastic collision, there is no loss in kinetic energy of the system. So, total kinetic energy is conserved. Also, the total momentum of the system is conserved.

Here, the nucleus is still. So, its initial kinetic energy is 0. So, the total initial kinetic energy will be equal to kinetic energy of the neutron only.

Now, final kinetic energy of the system will be equal to the initial kinetic energy.

Now, as the nucleus was at rest initially, so the final kinetic energy of the nucleus will be equal to the initial kinetic energy of the neutron.

Thus, all the kinetic energy of the neutron will be transferred to the nucleus and the neutron will come to rest after collision.

Therefore, the neutron loses all of its kinetic energy to nucleus.

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The wavelength of the visible line in the hydrogen spectrum that corresponds to m = 5 in the Balmer equation is: A. 656 nm. B. 4
BaLLatris [955]

Answer:

The wavelength of the visible line in the hydrogen spectrum is 434 nm.

Explanation:

It is given that, the wavelength of the visible line in the hydrogen spectrum that corresponds to n₂ = 5 in the Balmer equation.

For Balmer series, the wave number is given by :

\dfrac{1}{\lambda}=R(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})

R is the Rydberg's constant

For Balmer series, n₁ = 2. So,

\dfrac{1}{\lambda}=1.097\times 10^7\times (\dfrac{1}{2^2}-\dfrac{1}{5^2})

\lambda=4.34\times 10^{-7}\ m

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\lambda=434\ nm

So, the wavelength of the visible line in the hydrogen spectrum is 434 nm. Hence, this is the required solution.

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