\(\frac{E_e}{E_{\text{ph}}} = \frac{2c}{v}\)
\(\frac{E_e}{E_{\text{ph}}} = \frac{v}{2c}\)
\(\frac{p_e}{p_{\text{ph}}} = \frac{2c}{v}\)
\(\frac{p_e}{p_{\text{ph}}} = \frac{v}{2c}\)
\(\lambda_e = \lambda_{\text{ph}}\)
\(⇒\) \(\frac{h}{p_e} = \frac{hc}{E_{\text{ph}}}\)
\(⇒\) \(E_{\text{ph}} = p_e \cdot c = 2E_e \left(\frac{c}{v}\right)\)
\(⇒\) \(\frac{E_e}{E_{\text{ph}}} = \frac{v}{2c}\)
So, Option (B)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

The de Broglie wavelengths of a proton and an α particle are \( \lambda \) and \( 2\lambda \) respectively. The ratio of the velocities of proton and α particle will be:
The de-Broglie wavelength of an electron is the same as that of a photon. If the velocity of the electron is 25% of the velocity of light, then the ratio of the K.E. of the electron to the K.E. of the photon will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
The electron transport chain or system is the sequence of electron carriers, enzymes, and cytochrome that passes electrons from one to another through the redox reaction. It is electron transport-linked phosphorylation.
It contains flavin nucleotides (FAD), nicotinamide adenine dinucleotide (NAD), coenzyme Q, and cytochromes localized in F1 particles of mitochondria. It occurs in the inner mitochondrial membrane along with cristae.
In this process five (5) complexes are involved namely, I- NADH-UQ reductase, II- Succinate-UQ reductase, III- UQH2 -cytochrome C reductase, IV- Cytochrome C oxidase, and V is connected with F0−F1 particles.
In this process, NAD and FAD are minimized.
Steps: