The momentum of a photon can be calculated using the equation:
\(p = \frac{h}{\lambda}\)
Given:
\(\lambda = 2.2 \times 10^{-11} \, \text{m}\)
\(h = 6.6 \times 10^{-34} \, \text{J s}\)
Substituting the values into the equation, we have:
\(p = \frac{6.6 \times 10^{-34} \, \text{J s}}{2.2 \times 10^{-11} \, \text{m}}\)
\(p = \frac{6.6 \times 10^{-34} \, \text{Js}}{2.2 \times 10^{-11} \, \text{m}}\)
p ≈ 3 x 10^-23 Kg m/s
Therefore, the momentum of the photon is approximately (B) \(3 \times 10^{-23} \, \text{kg m/s}\)
Given below are two statements:
Given below are two statements:
In light of the above statements, choose the correct answer from the options given below:
The product (P) formed in the following reaction is:
In a multielectron atom, which of the following orbitals described by three quantum numbers will have the same energy in absence of electric and magnetic fields?
A. \( n = 1, l = 0, m_l = 0 \)
B. \( n = 2, l = 0, m_l = 0 \)
C. \( n = 2, l = 1, m_l = 1 \)
D. \( n = 3, l = 2, m_l = 1 \)
E. \( n = 3, l = 2, m_l = 0 \)
Choose the correct answer from the options given below: