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1 year ago

How many possible combinations are there for the values of l and ml when n = 2?

Physics

1 answer:

Alexeev081 [22]1 year ago

3 0

For n = 2, l will have two values 0,1

For l = 0, ml = 0

l = 1, ml = -1,0,1

These are the possible combinations.

The n is the principal quantum number

l is the azimuthal quantum number

ml is the magnetic quantum number.

The formula to find the azimuthal quantum number is

l = 0,1,2,3 .......n-1

If n = 2, the azimuthal quantum number,

l = 0 , 1

Then, the ml can be found using the formula,

ml = 2l +1

For l = 0 , ml = 2(0) + 1

= 1

Therefore, ml will have only one value, which is 0

For l =1 , ml = 2(1) + 1

= 3

Therefore, the ml will have three values, they are -1,0,1

Learn more about the quantum numbers in

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3 years ago

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Answer:

The total resistance of the wire is = $1.917\times10^{-3}$

Explanation:

Since the wires will both be in contact with the voltage source at the same time and the current flows along in their length-wise direction, the two wires will be considered to be in parallel.

Hence, for resistances in parallel, the total resistance, $R_{Total}$

$\frac{1}{R_{Total}} =\frac{1}{R_{cu} }+\frac{1}{R_{al}}$

Parameters given:

Length of wire = 1 m

Cross sectional area of copper $A_{cu}= \pi r^{2}= \pi \times (1\times 10^{-3} )^{2} =3.142\times10^{-6} m^{2}$

Cross sectional area of aluminium wire

$A_{al}= \pi( R^{2}-r^{2})\\\\ = \pi \times [ (2\times 10^{-3} )^{2}-(1\times 10^{-3} )^{2}] =9.42\times10^{-6} m^{2}\\$

Resistivity of copper $\rho _{cu}=1.7\times 10^{-8} \Omega .m$

Resistivity of Aluminium $\rho _{al}=2.8\times 10^{-8} \Omega .m$

Resistance of copper $R_{cu}= \frac{\rho_{cu} \times l}{A_{cu} } =\frac{1.7\times 10^{-8} \times 1}{3.142\times10^{-6} } =5.41\times 10^{-3}\Omega$

Resistance of aluminium $R_{al}= \frac{\rho_{al} \times l}{A_{al} } =\frac{2.8\times 10^{-8} \times 1}{9.42\times10^{-6} } =2.97\times 10^{-3}\Omega$

The total resistance of the wire can be obtained as follows;

$\frac{1}{R_{Total}} =\frac{1}{5.41\times10^{-3} }+\frac{1}{2.97\times10^{-3}}=521.52\frac{1}{\Omega}$

$R_{Total}= 1.917\times 10^{-3}\Omega$

∴ The total resistance of the wire = $1.917\times 10^{-3}\Omega$

4 0

3 years ago

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