यदि \(\lambda_{ o }\) और \(\lambda\) दहलीज़ी तरंगदैर्घ्य और आपतित प्रकाश का तरंग दैर्घ्य हों तो धातु स्थल से निकले प्रकाशीय इलेक्ट्रानों का वेग होगा।

A \(\sqrt {\frac{{2h}}{m}\left( {{\lambda _0} - \lambda } \right)} \)
B \(\sqrt {\frac{{2hc}}{m}\left( {{\lambda _0} - \lambda } \right)} \)
C \(\sqrt {\frac{{2hc}}{m}\left( {\frac{{{\lambda _0} - \lambda }}{{\lambda {\lambda _0}}}} \right)} \)
D \(\sqrt {\frac{{2h}}{m}\left( {\frac{1}{{{\lambda _0}}} - \frac{1}{\lambda }} \right)} \)

Verified Solution

Answer & Solution

Correct Answer

(C) \(\sqrt {\frac{{2hc}}{m}\left( {\frac{{{\lambda _0} - \lambda }}{{\lambda {\lambda _0}}}} \right)} \)

Step-by-step Solution

Detailed explanation

The kinetic enegry of the ejected electron is given by the equation \(hv = h{v_0} + \frac{1}{2}m{v^2}\,\,\,\because \,v = \frac{c}{\lambda }\) or \(\frac{{hc}}{\lambda } = \frac{{hc}}{{{\lambda _0}}} + \frac{1}{2}m{v^2}\)…

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.

Same subject

Explore more questions on app

From JEE Mains

Explore more questions on app