The angular part of the wave function for the electron in a hydrogen atom is proportional to sin2θcosθe2iφ. The values of the azimuthal quantum number (l) and the magnetic quantum number (m) are, respectivelya)2 and 2b)2 and –2c)3 and 2d)3 and –2Correct answer is option 'C'. Can you explain this answer?

Question

The angular part of the wave function for the electron in a hydrogen atom is proportional to sin2&theta;cos&theta;e2i&phi;. The values of the azimuthal quantum number (l) and the magnetic quantum number (m) are, respectivelya)2 and 2b)2 and &ndash;2c)3 and 2d)3 and &ndash;2Correct answer is option 'C'. Can you explain this answer?

Answer

(θ)cos(φ)

The angular part of the wave function for the electron in a hydrogen atom is given by the spherical harmonics:

Y(l,m)(θ,φ) = (-1)^m * sqrt((2l+1)/(4π) * (l-m)!/(l+m)!) * P(l,m)(cos(θ)) * exp(imφ)

Where:

- l is the principal quantum number (n) minus one, which determines the energy and size of the orbital.
- m is the magnetic quantum number, which determines the orientation of the orbital in space.
- θ is the polar angle, which measures the inclination of the electron's position vector with respect to the z-axis.
- φ is the azimuthal angle, which measures the angle of the electron's position vector with respect to the x-axis.

For l=1 (p orbital), m=±1, the spherical harmonics simplify to:

Y(1,1)(θ,φ) = -sqrt(3/(8π)) * sin(θ) * exp(iφ)

Y(1,-1)(θ,φ) = sqrt(3/(8π)) * sin(θ) * exp(-iφ)

Y(1,0)(θ,φ) = sqrt(3/(4π)) * cos(θ)

The angular part of the wave function for the electron in a hydrogen atom is the linear combination of these three spherical harmonics, weighted by the corresponding coefficient (c1, c-1, c0) that depends on the quantum numbers (n, l, m) and the normalization condition:

Ψ(θ,φ) = c1 * Y(1,1)(θ,φ) + c-1 * Y(1,-1)(θ,φ) + c0 * Y(1,0)(θ,φ)

To simplify this expression, we can use the trigonometric identity sin2(θ) = (1-cos2(θ)) and the property that exp(iφ)+exp(-iφ)=2cos(φ):

Ψ(θ,φ) = sqrt(3/(4π)) * [c1 * (-exp(iφ)*sin(θ)) + c-1 * (exp(-iφ)*sin(θ)) + c0 * cos(θ)]

Ψ(θ,φ) = sqrt(3/(4π)) * sin(θ) * [c1 * (-exp(iφ)) + c-1 * (exp(-iφ)) + c0 * cos(θ)/sin(θ)]

We can define the constants A and B as:

A = (c1-c-1)/sqrt(2)

B = i(c1+c-1)/sqrt(2)

Then, we can rewrite the angular part of the wave function as:

Ψ(θ,φ) = A * (-exp(iφ)*sin(θ)) + B * (exp(-iφ)*sin(θ)) + c0 * cos(θ)

Ψ(θ,φ) = sqrt(3/(4π)) * sin(θ) * [A * (-exp(iφ)) + B * (exp(-iφ)) + c0 * cos(θ)/sin(θ)]

Finally, if we choose the phases of A and B such that:

A = -sin(α)

B = cos(α)

Then, the

Answer

We get value of m from.esponential part and vlue of l get from maximum.power of cos and sin

Similar Chemistry Doubts

Which of the following is true for the radial part of the H atom wave function Rn, l(r) (n-principal quantum number and l-azimuthal quantum number) and the nodes associated with them

For an electron in a hydrogen atom, the wave function, ψ is proportional to exp(-r/a0), where a0 is the Bohr’s radius. What is the ratio of the probability of finding the electron at the nucleus to the probability of finding it at a0

Evaluate the root mean square distance, (r2)1/2of the electron from the nucleus in the hydrogenatom described by the wave function[in pm rounded up to two decimal place] Correct answer is between '91.00,92.00'. Can you explain this answer?

The wave function for an electron in a hydrogen atom is sai =Ne^-r/a. Here r is the distance from the distance from nucleus to the electron and a is the bones radius. Calculate naormailsation factor,N.?

A higher energy wave function of the electron in a nanotube is described by the normalized wave function (2/L)1/2sin(2πx/L). What is the percentage probability of finding the electron between x = L/4 and x = L/2? Correct answer is '25'. Can you explain this answer?

Top Courses for Chemistry

Suggested Free Tests

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams