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2 years ago

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

Physics

1 answer:

Alexeev081 [22]2 years 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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Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m

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The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

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Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = $0.407\times 10^{-3}\ m$

Energy stored in the capacitor (U) = $7.87\ nJ=7.87\times 10^{-9}\ J$

Assuming the medium to be air.

So, permittivity of space (ε) = $8.854\times 10^{-12}\ F/m$

Area of the square plates is given as:

$A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2$

Capacitance of the capacitor is given as:

$C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F$

Now, we know that, the energy stored in a parallel plate capacitor is given as:

$U=\frac{CE^2d^2}{2}$

Rewriting in terms of 'E', we get:

$E=\sqrt{\frac{2U}{Cd^2}}$

Now, plug in the given values and solve for 'E'. This gives,

$E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C$

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

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