બે સમાન ઘાતુની પ્લેટ $A$ અને $ B$ પર ${\lambda _A}$અને${\lambda _B} ({\lambda _A} = 2{\lambda _B})$ તરંગલંબાઇ આપાત કરતાં ફોટો-ઇલેકટ્રોનની ગતિઊર્જા કેટલી થાય?

Q: બે સમાન ઘાતુની પ્લેટ $A$ અને $ B$ પર ${\lambda _A}$અને${\lambda _B} ({\lambda _A} = 2{\lambda _B})$ તરંગલંબાઇ આપાત કરતાં ફોટો-ઇલેકટ્રોનની ગતિઊર્જા કેટલી થાય?

b (b) $\frac{{hc}}{\lambda } = {W_0} + {K_{\max }}$ $ \Rightarrow \frac{{hc}}{{{\lambda _A}}} = {W_0} + {K_A}$ ..$.(i)$ and $\frac{{hc}}{{{\lambda _B}}} = {W_0} + {K_B}$ ...$(ii)$ Subtracting $(i)$ from $(ii)$, $hc\left[ {\frac{1}{{{\lambda _B}}} - \frac{1}{{{\lambda _A}}}} \right] = {K_B} - {K_A}$ ==> $hc\left[ {\frac{1}{{{\lambda _B}}} - \frac{1}{{2{\lambda _B}}}} \right] = {K_B} - {K_A}$ ==>$\frac{{hc}}{{2{\lambda _B}}} = {K_B} - {K_A}$ ...$(iii)$ From $(ii)$ and $(iii)$, $2{K_B} - 2{K_A} = {W_0} + {K_B}$ $ \Rightarrow {K_B} - 2{K_A} = {W_0}$ $ \Rightarrow {K_A} = \frac{{{K_B}}}{2} - \frac{{{W_0}}}{2}$ which gives ${K_A} < \frac{{{K_B}}}{2}$.

Question

A$2{K_A} = {K_B}$
B${K_A} < {K_B}/2$
C${K_A} = 2{K_B}$
D${K_A} = {K_B}/2$

Diffcult

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Solution

b
(b) $\frac{{hc}}{\lambda } = {W_0} + {K_{\max }}$

$ \Rightarrow \frac{{hc}}{{{\lambda _A}}} = {W_0} + {K_A}$ ..$.(i)$ and $\frac{{hc}}{{{\lambda _B}}} = {W_0} + {K_B}$ ...$(ii)$

Subtracting $(i)$ from $(ii)$, $hc\left[ {\frac{1}{{{\lambda _B}}} - \frac{1}{{{\lambda _A}}}} \right] = {K_B} - {K_A}$

==> $hc\left[ {\frac{1}{{{\lambda _B}}} - \frac{1}{{2{\lambda _B}}}} \right] = {K_B} - {K_A}$

==>$\frac{{hc}}{{2{\lambda _B}}} = {K_B} - {K_A}$ ...$(iii)$ From $(ii)$ and $(iii)$, $2{K_B} - 2{K_A} = {W_0} + {K_B}$

$ \Rightarrow {K_B} - 2{K_A} = {W_0}$

$ \Rightarrow {K_A} = \frac{{{K_B}}}{2} - \frac{{{W_0}}}{2}$ which gives ${K_A} < \frac{{{K_B}}}{2}$.

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