Question

For an element with atomic number 29 write the value of n and l for its electron in its valence shell

Answer 1

Answer:

The value of n is $4$.

The value of l is $0$.

Explanation:

Given data,

The atomic number of the given element = $29$

As we know,

The element with an atomic number $29$ is copper ( Cu ).

The value of n and l in its valence shell =?

Firstly, we have to write the electronic configuration of the element copper with an atomic number $29$.

• Electronic configuration = $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{1} 3d^{10}$
• The valence shell is 4s.

As we know,

• n = principal quantum number
• It tells about the principal energy shell to which the electron belongs.

Therefore,

• n = 4.

As we know,

• l = Azimuthal quantum number
• It tells about the orbital angular momentum and the shape of the orbitals.

For s, l = 0,

For p, l = 1

For d, l = 2

For f, l = 3

As the valence shell is 4s; therefore, the value of l is $0$.