Question:
The spatial component of a wave function is given by
\(\psi = A[\sin(ax)\cos(bx) + \cos(ax)\sin(bx)]\). The wavelength of the wave is:
The spatial component of a wave function is given by
\(\psi = A[\sin(ax)\cos(bx) + \cos(ax)\sin(bx)]\). The wavelength of the wave is:
Show Hint
Always simplify wave expressions using trig identities before identifying wave number.
Updated On: Jun 20, 2026
- \(\frac{1}{a+b}\)
- \(a+b\)
- \(\frac{2\pi}{a+b}\)
- \(2\pi\)
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The Correct Option is C
Solution and Explanation
Step 1: Use trigonometric identity.
\[ \sin(ax)\cos(bx) + \cos(ax)\sin(bx) = \sin((a+b)x) \] So wave function becomes: \[ \psi = A\sin((a+b)x) \]
Step 2: Identify wave number.
Standard form: \[ \psi = A\sin(kx) \] So: \[ k = a + b \]
Step 3: Relation between wavelength and wave number.
\[ \lambda = \frac{2\pi}{k} \]
Step 4: Substitute value.
\[ \lambda = \frac{2\pi}{a+b} \]
Step 5: Final conclusion.
\[ \boxed{\frac{2\pi}{a+b}} \]
\[ \sin(ax)\cos(bx) + \cos(ax)\sin(bx) = \sin((a+b)x) \] So wave function becomes: \[ \psi = A\sin((a+b)x) \]
Step 2: Identify wave number.
Standard form: \[ \psi = A\sin(kx) \] So: \[ k = a + b \]
Step 3: Relation between wavelength and wave number.
\[ \lambda = \frac{2\pi}{k} \]
Step 4: Substitute value.
\[ \lambda = \frac{2\pi}{a+b} \]
Step 5: Final conclusion.
\[ \boxed{\frac{2\pi}{a+b}} \]
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