I can’t figure it out! <br> Please help I have a test tomorrow

An electron in a mercury atom jumps from level a to level g by absorbing a single

statuscvo [17]

Answer:

7.39ev

Explanation:

Energy levels are found inside the atom. Electrons occupy these energy levels depending on the energy they possess. Electrons can move from one energy level to another due to absorption or emission of a photon or other factors. As the electron, jumps from a higher energy level to a lower energy level emitting a photon of measurable frequency, the photon carries energy equal to the amount of energy between the gap of the levels. This idea was first proposed by Neils Bohr and became the forerunner of the wave mechanical model of the atom.

Hence the energy of a photon is the energy of the gap between the two energy levels. Since Ea= 2.48ev and Eg= 10.38 ev.

If an electron jumps from Ea to Eg, the energy of the photon absorbed is given by;

E=Eg-Ea

E= 10.38ev - 2.48ev

E= 7.39ev

3 0

3 years ago

Choose the more reactive metal. Al or Zn

horsena [70]

The truth is both of them are more reactive metal.

8 0

3 years ago

Read 2 more answers

What are the two distinct ways in which energy moves outward from the solar core to photosphere?

butalik [34]

Answer:

The energy may be carried in the form of (1) radiation, where energy travels in the form of light, and (2) convection, where energy is carried by physical motion of upwelling solar gas.

Explanation:

6 0

2 years ago

What gravitational force does the moon produce on the earth is their centers are 3. 88x10^8 m apart and the moon has a mass of 7

vitfil [10]

The Moon is 3.8 108 m from Earth and has a mass of 7.34 1022 kg. 5.97 1024 kg is the mass of the Earth.

<h3>What kind of gravitational pull does the moon have on the planet?</h3>

On the surface of the Moon, the acceleration caused by gravity around 1.625 m/s2 which is 16.6% greater than on the surface of the Earth 0.166.

<h3>What does the Earth's center's gravitational pull feel like?</h3>

Gravity is zero if you are in the centre of the earth since everything around you is pulling "up" (up is the only direction).

<h3>Where is the Earth's and the moon's gravitational centre?</h3>

It is around 1700 kilometres below Earth's surface.

To know more about gravitational force visit:-

brainly.com/question/12528243

#SPJ4

6 0

1 year ago

A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

6 0

3 years ago

I can’t figure it out! Please help I have a test tomorrow