PYQ NEET- ଅଣ୍ଡାକର ଗଠନ ଲାଇନ୍ଜ୍-5

ପ୍ରଶ୍ନ: ଯେତେବେଳେ ତରଙ୍ଗବଳା ବାଣୀ $300 \mathrm{~nm}$ ଉପରରେ ଏକ ଧାତୁର ପୃଷ୍ଠରେ ପଡ଼ିଯିବ, ତାହା $1.68 \times 10^5 \mathrm{~J}$ $\mathrm{mol}^{-1}$ କିନେଟିକ୍ ଷଟିନ୍ ବିକିରଣ କରିଥାଏ। ଧାତୁର ପରେମାନଙ୍କରୁ ଏକ ବିକ୍ରିୟାକାରୀ ବାଣିକୀକୁ ଦୂର କରିବା ପାଇଁ ଆବଶ୍ୟକୀୟ ଶକ୍ତି କିଏ?#

$$ \left(\mathrm{h}=6.626 \times 10^{-34} \mathrm{Js}, \mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}, \mathrm{~N}_{\mathrm{A}}=6.022 \times 10^{23} \mathrm{~mol}^{-1}\right) $$

A) $2.31 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1}$

B) $2.31 \times 10^6 \mathrm{~J} \mathrm{~mol}^{-1}$

C) $3.84 \times 10^4 \mathrm{~J} \mathrm{~mol}^{-1}$

D) $3.84 \times 10^{-19} \mathrm{~J} \mathrm{~mol}^{-1}$

ଉତ୍ତର: $2.31 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1}$#

ସମାଧାନ:#

ଏକ ଫଟନର ଶକ୍ତି $=\frac{h c}{\lambda}(\lambda=300 \mathrm{~nm})$
ଏକ ମୋଲ ଫଟନର ପାଇଁ, $E=\frac{h c}{\lambda} \times N_A$
$$ \begin{aligned} & E=\frac{6.626 \times 10^{-34} \times 3 \times 10^8 \times 6.023 \times 10^{23}}{300 \times 10^{-9}} \ & E=3.99 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1} \ & \text { Kinetic energy }=1.68 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1} \ & W_0=E-K . E \ & =3.99 \times 10^5-1.68 \times 10^5 \ & =2.31 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1} \end{aligned} $$