TS EAMCET · Physics · Atomic Physics

The light emitted in the transition $n = 3$ to $n = 2$ (where $n$ is the principal quantum number of the state) in hydrogen is called $H_{α}$ -light. Find the maximum work function that a metal can have so that $H_{α}$ -light can emit photoelectrons from it.

A $1.5 eV$
B $2.89 eV$
C $1.89 eV$
D $3.5 eV$

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Answer & Solution

Correct Answer

(C) $1.89 eV$

Step-by-step Solution

Detailed explanation

Energy of Hα-light =-13.6132-122 eV=-13.619-14 eV≈1.89 eV Maximum work function of the metal should be equal to this value for it to emit photoelectrons from it.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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The light emitted in the transition n=3 to n=2 (where n is the principal quantum number of the state) in hydrogen is called Hα-light. Find the maximum work function that a metal can have so that Hα-light can emit photoelectrons from it.

Step-by-step Solution

The light emitted in the transition $n = 3$ to $n = 2$ (where $n$ is the principal quantum number of the state) in hydrogen is called $H_{α}$ -light. Find the maximum work function that a metal can have so that $H_{α}$ -light can emit photoelectrons from it.