Given Formula:
The energy of a photon is given by the formula:
\[ E = \frac{1240}{\lambda (\text{nm})} \, \text{eV} \]
Step 1: Substitute the wavelength:
Substitute the wavelength into the equation:
\[ E = \frac{1240}{242} \, \text{eV} \]
Step 2: Simplify to find the energy in eV:
After performing the calculation:
\[ E = 5.12 \, \text{eV} \]
Step 3: Convert to Joules per atom:
To convert from eV to Joules, multiply by \( 1.6 \times 10^{-19} \, \text{J/eV} \):
\[ 5.12 \times 1.6 \times 10^{-19} = 8.198 \times 10^{-19} \, \text{J/atom} \]
Step 4: Convert to kJ/mol:
To convert from Joules per atom to kJ per mole, multiply by Avogadro's number (\( 6.022 \times 10^{23} \)) and divide by 1000:
\[ 8.198 \times 10^{-19} \times 6.022 \times 10^{23} = 494 \, \text{kJ/mol} \]
Final Answer:
The energy is \( 494 \, \text{kJ/mol} \).
\[ E = \frac{1240}{\lambda (\text{nm})} \, \text{eV} \]
\[ E = \frac{1240}{242} \, \text{eV} \]
\[ E = 5.12 \, \text{eV} \]
\[ E = 5.12 \times 1.6 \times 10^{-19} \, \text{J/atom} \]
\[ E = 8.198 \times 10^{-19} \, \text{J/atom} \]
\[ E = 494 \, \text{kJ/mol} \]
Which of the following is the correct electronic configuration for \( \text{Oxygen (O)} \)?
Which of the following is/are correct with respect to the energy of atomic orbitals of a hydrogen atom?
(A) \( 1s<2s<2p<3d<4s \)
(B) \( 1s<2s = 2p<3s = 3p \)
(C) \( 1s<2s<2p<3s<3p \)
(D) \( 1s<2s<4s<3d \)
Choose the correct answer from the options given below:
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).

In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by: