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Wittaler [7]
3 years ago
5

Why does the emission spectrum of hydrogen or unknown source have a distinct line spectrum instead of a continuous one?

Physics
1 answer:
aleksandr82 [10.1K]3 years ago
3 0

Answer

When an electron makes transition from a state of higher energy to a state of lower energy it does so by emitting energy in form of radiation in the visible spectrum of light.

Since the basic postulates of the atomic theory is that the energy that the electron possess in it's orbit's takes only discrete values and cannot take any random value thus when an electron makes a transition from a state of higher energy to state of lower energy it will emit radiation with energy equal to difference between the energy levels of the 2 orbit's thus we only observe discrete lines.

Mathematically when an electron makes a transition between states the wavelength of light it releases is given by

\frac{1}{\lambda }=R_{\infty }(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}})

where R_{\infty } is Rydberg constant

n_{1} is upper energy level

n_{2} is lower energy level

thus we can see that only discrete wavelength's are released and not continuous wavelength's of light.

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The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is sudde
Afina-wow [57]

Answer:Resistor-Capacitor (RC) circuits, when driven by a voltage/current source, display a type of time-dependent charging or discharging since the charge from the capacitor goes through the resistor. When considering a discharging phase, the time-dependent voltage takes the form

Explanation:

Vt = Voe^-t/RC

Vt equal time dependent voltage

R is the resistance

C is the capacitance of the oscilloscope

Given

Area of capacitor equal to 0.1 * 0.002m

Distance between plates equal 1/1000 = 0.0001m

Vo = 100v supply voltage

Resistance R =:1000ohms

Vt time dependent voltage = 50V

To find capacitance C = Eo(A/d)

C = (8.85 * 10^-12)* (0.002/0.001)

C= 1.77 * 10^-11

Solving the equation Vt = Voe^-t/RC

t = 0.6931RC

t = 0.6931(1000)(1.77*10^-11)

t = 1.2*10^-9

t = 1.2ns

It will take the oscilloscope 1.2ns to reach 50V

8 0
3 years ago
According to the periodic table, the average atomic mass of helium is A) 2 amu. B) 2.0026 amu. C) 4.0026 amu. D) 6.0026 amu.
RUDIKE [14]

Answer:

C) 4.0026 amu

Explanation:

Helium (He) is the 2nd element on the Periodic Table.

It's neutral atom has 2 protons and 2 electrons.

It is in the 1st period and the 18th row.

It is also a Noble gas.

3 0
3 years ago
Pls help I will mark brainliest
earnstyle [38]
It will be unstable system because it will not be able to recover from the disturbance
7 0
3 years ago
Read 2 more answers
3) Principles of rectilinear proportion of light 9​
8_murik_8 [283]

Answer:Principle of rectilinear propagation of light

Explanation:Principle of rectilinear propagation of light

Rectilinear propagation of light refers to the propensity of light to travel along a straight line without any interference in its trajectory. ... It is because light travels along a straight line and leaves only the areas where the object interferes.

7 0
2 years ago
A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri
GREYUIT [131]

Answer:

|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}

Now, substitute this into 'dF':

d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)

Now we can integrate dF over the rod.

\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)

4 0
3 years ago
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