Step 1: Use the relation between energy and wavelength.
The energy of a photon is related to its wavelength by: \[ E = \frac{hc}{\lambda} \Rightarrow \lambda = \frac{hc}{E} \]
Where:
Step 2: Calculate the wavelength.
\[ \lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3 \times 1.6 \times 10^{-19}} = \frac{1.9878 \times 10^{-25}}{4.8 \times 10^{-19}} \approx 4.14 \times 10^{-7} \, \text{m} = 414 \, \text{nm} \] Step 3: Select the closest option.
The wavelength is approximately 413 nm, which matches option (3).
If \( \vec{u}, \vec{v}, \vec{w} \) are non-coplanar vectors and \( p, q \) are real numbers, then the equality:
\[ [3\vec{u} \quad p\vec{v} \quad p\vec{w}] - [p\vec{v} \quad \vec{w} \quad q\vec{u}] - [2\vec{w} \quad q\vec{v} \quad q\vec{u}] = 0 \]
holds for:
Statement-I: In the interval \( [0, 2\pi] \), the number of common solutions of the equations
\[ 2\sin^2\theta - \cos 2\theta = 0 \]
and
\[ 2\cos^2\theta - 3\sin\theta = 0 \]
is two.
Statement-II: The number of solutions of
\[ 2\cos^2\theta - 3\sin\theta = 0 \]
in \( [0, \pi] \) is two.
If \( A \) and \( B \) are acute angles satisfying
\[ 3\cos^2 A + 2\cos^2 B = 4 \]
and
\[ \frac{3 \sin A}{\sin B} = \frac{2 \cos B}{\cos A}, \]
Then \( A + 2B = \ ? \)
If
\[ \sin \theta + 2 \cos \theta = 1 \]
and
\[ \theta \text{ lies in the 4\textsuperscript{th} quadrant (not on coordinate axes), then } 7 \cos \theta + 6 \sin \theta =\ ? \]