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Setler [38]
3 years ago
10

What is meant by line spectrum? Explain how line spectrum can be used for the identification of elements?

Physics
1 answer:
Harlamova29_29 [7]3 years ago
8 0
The line spectrum is also known as the emission light spectrum.
The ELS shows what colors of light are released when an element(or a specific compoud) is burned or electricity is sent through the gaseous form.

The ELS can be used to identify elements because each element has a different ELS. Elements like Tungsten produce a white light when burned, so their ELS has many line close together. This is why tungsten was used for lightbulbs.
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A test charge of +4 µC is placed halfway between a charge of +6 µC and another of +2 µC separated by 20 cm. (a) What is the magn
S_A_V [24]

Answer:

(a) Magnitude: 14.4 N

(b) Away from the +6 µC charge

Explanation:

As the test charge has the same sign, the force that the other charges exert on it will be a repulsive force. The magnitude of each of the forces will be:

F_e = K\frac{qq_{test}}{r^2}

K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q and qtest is the charge of the particles, and r is the distance between the particles.

Let's say that a force that goes toward the +6 µC charge is positive, then:

F_e_1 = K\frac{q_1q_{test}}{r^2}=-9*10^9 \frac{Nm^2}{C^2} \frac{6*10^{-6}C*4*10^{-6}C}{(0.1m)^2} =-21.6 N

F_e_2 = K\frac{q_2q_{test}}{r^2}=9*10^9 \frac{Nm^2}{C^2} \frac{2*10^{-6}C*4*10^{-6}C}{(0.1m)^2} =7.2 N

The magnitude will be:

F_e = -21.6 + 7.2 = -14.4 N, away from the +6 µC charge

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Consider a model of a hydrogen atom in which an electron is in a circular orbit of radius r = 5.92×10−11 m around a stationary p
DaniilM [7]

Answer:

2.068 x 10^6 m / s

Explanation:

radius, r = 5.92 x 10^-11 m

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

As the electron is revolving in a circular path, it experiences a centripetal force which is balanced by the electrostatic force between the electron and the nucleus.

centripetal force = \frac{mv^{2}}{r}

Electrostatic force = \frac{kq^{2}}{r^{2}}

where, k be the Coulombic constant, k = 9 x 10^9 Nm^2 / C^2

So, balancing both the forces we get

\frac{kq^{2}}{r^{2}}=\frac{mv^{2}}{r}

v=\sqrt{\frac{kq^{2}}{mr}}

v=\sqrt{\frac{9\times 10^{9}\times1.6\times 10^{-19}\times 1.6\times 10^{-19}}{9.1\times 10^{-31}\times 5.92\times10^{-11}}}

v = 2.068 x 10^6 m / s

Thus, the speed of the electron is give by  2.068 x 10^6 m / s.

6 0
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