एक प्रकाश उत्सर्जक सेल में कार्यकारी तरंगदैर् | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$h
u  - {W_0} = \frac{1}{2}mv_{\max }^2$

$\Rightarrow \frac{{hc}}{\lambda } - \frac{{hc}}{{{\lambda _0}}} = \frac{1}{2}mv_{\max }^2$

$ \Ri

Vedclass · Accepted Answer

$ > v\;{(4/3)^{1/2}}$

Vedclass · Answer

$h
u  - {W_0} = \frac{1}{2}mv_{\max }^2$

$\Rightarrow \frac{{hc}}{\lambda } - \frac{{hc}}{{{\lambda _0}}} = \frac{1}{2}mv_{\max }^2$

$ \Rightarrow hc\left( {\frac{{{\lambda _0} - \lambda }}{{\lambda {\lambda _0}}}} ight) = \frac{1}{2}mv_{\max }^2$

$ \Rightarrow {v_{\max }} = \sqrt {\frac{{2hc}}{m}\left( {\frac{{{\lambda _0} - \lambda }}{{\lambda {\lambda _0}}}} ight)} $

जब तरंगदैध्र्य $\lambda $ एवं वेग $v$ है, तब

$v = \sqrt {\frac{{2hc}}{m}\left( {\frac{{{\lambda _0} - \lambda }}{{\lambda {\lambda _0}}}} ight)} $                                               &hellip;. $(i)$

 जब तरंगदैध्र्य $\frac{{3\lambda }}{4}$ एवं वेग $v$' है तब

$v' = \sqrt {\frac{{2hc}}{m}\left[ {\frac{{{\lambda _0} - (3\lambda /4)}}{{(3\lambda /4) 	imes {\lambda _0}}}} ight]} $                           &hellip;.$(ii)$

 समीकरण $(ii)$ को $(i)$ से भाग देने पर

 $\frac{{v'}}{v} = \sqrt {\frac{{[{\lambda _0} - (3\lambda /4)]}}{{\frac{3}{4}\lambda {\lambda _0}}} 	imes \frac{{\lambda {\lambda _0}}}{{{\lambda _0} - \lambda }}} $

 $v' = v{\left( {\frac{4}{3}} ight)^{1/2}}\sqrt {\frac{{[{\lambda _0} - (3\lambda /4)]}}{{{\lambda _0} - \lambda }}} $ अर्थात् $v' > v{\left( {\frac{4}{3}} ight)^{1/2}}$

Similar Questions