The quantum mechanical model of the hydrogen atom requires that if the angular momentum number is 4, the number of different permitted orbital magnetic quantum numbers will be:
nine
The quantum mechanical model describes the behavior of electrons in atoms. This model uses a set of four quantum numbers to define the unique state of an electron. These numbers include the principal quantum number (n), the angular momentum or azimuthal quantum number (l), the magnetic quantum number ($m_l$), and the spin quantum number ($m_s$). Understanding the relationship between these numbers is crucial for determining electron properties.
The angular momentum number, denoted by 'l', describes the shape of an electron's orbital and the orbital angular momentum. Its values depend on the principal quantum number (n), ranging from 0 to n-1. Each 'l' value corresponds to a specific subshell: l=0 is an s orbital, l=1 is a p orbital, l=2 is a d orbital, and so on.
In this question, we are given that the angular momentum number is 4. This means we are considering an orbital with $l = 4$.
The orbital magnetic quantum number, denoted by '$m_l$', describes the orientation of an orbital in space. For a given angular momentum number 'l', the possible values of '$m_l$' are integers ranging from $-l$ to $+l$, including zero.
The number of different permitted orbital magnetic quantum numbers for a given 'l' is determined by the formula:
Number of $m_l$ values $= 2l + 1$
Given the angular momentum number $l = 4$, we can calculate the number of different permitted orbital magnetic quantum numbers using the formula:
Number of $m_l$ values $= 2l + 1$
Substitute the given value of $l$ into the formula:
Number of $m_l$ values $= 2(4) + 1$
Number of $m_l$ values $= 8 + 1$
Number of $m_l$ values $= 9$
The permitted values for $m_l$ when $l=4$ are:
Counting these values, we find there are exactly nine different permitted orbital magnetic quantum numbers.
Based on the calculations, if the angular momentum number for a hydrogen atom is 4, the number of different permitted orbital magnetic quantum numbers will be nine.
Which of the following was not observed by Rutherford using the scattering of α-rays?
1. Most of the α-particles get slightly deflected from their path.
2. Fewer α-particles get deflected at greater angles.
After completing the gold foil experiment, Rutherford concluded that the size of the nucleus is very small compared to the size of the atom. This is because:
The nucleus of an atom was discovered by:
_______ specifies the preferred orientation in the orbital space of the given energy and size.
What is the ratio of total kinetic energies in laboratory system (TL) and centre of mass system (TC ) in the scattering with projectile of mass m1 and target of mass m2?