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Question Number 22797 by Tinkutara last updated on 22/Oct/17
He^+ in ground state is hit by photon thus exciting electron to second excited state. The energy absorbed in this process is (approx) (1) 2.18 × 10^(−14) J (2) 7.75 × 10^(−18) J (3) 4.5 × 10^(−18) J (4) 5 × 10^(−18) J
$$\mathrm{He}^{+} \:\mathrm{in}\:\mathrm{ground}\:\mathrm{state}\:\mathrm{is}\:\mathrm{hit}\:\mathrm{by}\:\mathrm{photon} \\ $$$$\mathrm{thus}\:\mathrm{exciting}\:\mathrm{electron}\:\mathrm{to}\:\mathrm{second}\:\mathrm{excited} \\ $$$$\mathrm{state}.\:\mathrm{The}\:\mathrm{energy}\:\mathrm{absorbed}\:\mathrm{in}\:\mathrm{this} \\ $$$$\mathrm{process}\:\mathrm{is}\:\left(\mathrm{approx}\right) \\ $$$$\left(\mathrm{1}\right)\:\mathrm{2}.\mathrm{18}\:×\:\mathrm{10}^{−\mathrm{14}} \:\mathrm{J} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{7}.\mathrm{75}\:×\:\mathrm{10}^{−\mathrm{18}} \:\mathrm{J} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{4}.\mathrm{5}\:×\:\mathrm{10}^{−\mathrm{18}} \:\mathrm{J} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{5}\:×\:\mathrm{10}^{−\mathrm{18}} \:\mathrm{J} \\ $$
Answered by = last updated on 22/Oct/17
using balmer′formula : E_(required) = 2.18 ×10^(−18) ×Z^(2 ) {(1/n_1 ^2 ) +(1/n_2 ^2 )} J ⇒ E = 2.18 × 10^(−18) ×2^2 {(1/1)+(1/3^2 )} ⇒E = 7.75×10^(−18) J
$$ \\ $$$${using}\:{balmer}'{formula}\:: \\ $$$${E}_{{required}} \:=\:\mathrm{2}.\mathrm{18}\:×\mathrm{10}^{−\mathrm{18}} ×{Z}^{\mathrm{2}\:} \left\{\frac{\mathrm{1}}{{n}_{\mathrm{1}} ^{\mathrm{2}} }\:+\frac{\mathrm{1}}{{n}_{\mathrm{2}} ^{\mathrm{2}} }\right\}\:{J} \\ $$$$\Rightarrow\:{E}\:=\:\mathrm{2}.\mathrm{18}\:×\:\mathrm{10}^{−\mathrm{18}} ×\mathrm{2}^{\mathrm{2}} \left\{\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\right\} \\ $$$$\Rightarrow{E}\:=\:\mathrm{7}.\mathrm{75}×\mathrm{10}^{−\mathrm{18}} {J} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Tinkutara last updated on 22/Oct/17
But the formula is E = 2.18 ×10^(−18) Z^(2 ) ((1/n_1 ^2 )−(1/n_2 ^2 ))J. Why you have written + there?
$$\mathrm{But}\:\mathrm{the}\:\mathrm{formula}\:\mathrm{is} \\ $$$${E}\:=\:\mathrm{2}.\mathrm{18}\:×\mathrm{10}^{−\mathrm{18}} {Z}^{\mathrm{2}\:} \left(\frac{\mathrm{1}}{{n}_{\mathrm{1}} ^{\mathrm{2}} }−\frac{\mathrm{1}}{{n}_{\mathrm{2}} ^{\mathrm{2}} }\right)\mathrm{J}.\:\mathrm{Why} \\ $$$$\mathrm{you}\:\mathrm{have}\:\mathrm{written}\:+\:\mathrm{there}? \\ $$
Commented by = last updated on 22/Oct/17
sorry i meant − sign only and not +
$${sorry}\:{i}\:{meant}\:−\:{sign}\:{only}\:\:{and}\:{not}\:+ \\ $$
Commented by Tinkutara last updated on 22/Oct/17
Then answer would not change?
$$\mathrm{Then}\:\mathrm{answer}\:\mathrm{would}\:\mathrm{not}\:\mathrm{change}? \\ $$
Commented by À. last updated on 22/Oct/17
the calculations have been done according to negative sign
$${the}\:{calculations}\:{have}\:{been}\:{done} \\ $$$${according}\:{to}\:{negative}\:{sign} \\ $$
Commented by Tinkutara last updated on 22/Oct/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
Commented by Rasheed.Sindhi last updated on 22/Oct/17
To the person who gave above answer. What is your id? What id can be thought by “Answer: by on 22/10/17 ”
$$\mathrm{To}\:\mathrm{the}\:\mathrm{person}\:\mathrm{who}\:\mathrm{gave}\:\mathrm{above} \\ $$$$\mathrm{answer}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{your}\:\mathrm{id}? \\ $$$$\mathrm{What}\:\mathrm{id}\:\mathrm{can}\:\mathrm{be}\:\mathrm{thought}\:\mathrm{by} \\ $$$$“\mathrm{Answer}:\:\mathrm{by}\:\mathrm{on}\:\mathrm{22}/\mathrm{10}/\mathrm{17}\:'' \\ $$
Commented by Physics lover last updated on 22/Oct/17
lol,i am the person who gave the answer i selected an emoji as my username at first the server accepted but it dint show that emoji here,instead it showed a blank space.
$${lol},{i}\:{am}\:{the}\:{person}\:{who}\:{gave}\:{the}\:{answer} \\ $$$${i}\:{selected}\:{an}\:{emoji}\:{as}\:{my}\:{username} \\ $$$${at}\:{first}\:{the}\:{server}\:{accepted}\:{but} \\ $$$${it}\:{dint}\:{show}\:{that}\:{emoji}\:{here},{instead} \\ $$$${it}\:{showed}\:{a}\:{blank}\:{space}. \\ $$
Commented by Physics lover last updated on 23/Oct/17
thats why you see a space there− answer by“ ” on 22/10/17
$${thats}\:{why}\:{you}\:{see}\:{a}\:{space}\:{there}− \\ $$$${answer}\:{by}“\:\:''\:{on}\:\mathrm{22}/\mathrm{10}/\mathrm{17} \\ $$
Commented by Rasheed.Sindhi last updated on 23/Oct/17
So, in addition to physics you are now experimenting on id! hahaha Anyway thanx to explain.... But there isn′t even a blank space between “Answer: by”and “on 22/10/17”!
$$\mathrm{So},\:\mathrm{in}\:\mathrm{addition}\:\mathrm{to}\:\mathrm{physics}\:\mathrm{you}\:\mathrm{are} \\ $$$$\mathrm{now}\:\mathrm{experimenting}\:\mathrm{on}\:\mathrm{id}!\:\mathrm{hahaha} \\ $$$$\mathrm{Anyway}\:\mathrm{thanx}\:\mathrm{to}\:\mathrm{explain}…. \\ $$$$\mathrm{But}\:\mathrm{there}\:\mathrm{isn}'\mathrm{t}\:\mathrm{even}\:\mathrm{a}\:\mathrm{blank}\:\mathrm{space}\:\mathrm{between} \\ $$$$“\mathrm{Answer}:\:\mathrm{by}''\mathrm{and}\:“\mathrm{on}\:\mathrm{22}/\mathrm{10}/\mathrm{17}''! \\ $$
Commented by Physics lover last updated on 23/Oct/17
well,there is one space there.lol
$${well},{there}\:{is}\:{one}\:{space}\:{there}.{lol} \\ $$
Commented by Physics lover last updated on 23/Oct/17
you would also see a question mark symbol in the next . i used the username “π.”there It displayed question mark instead of π and showed the dot as it is
$${you}\:{would}\:{also}\:{see}\:{a}\:{question}\: \\ $$$${mark}\:{symbol}\:{in}\:{the}\:{next}\:. \\ $$$${i}\:{used}\:{the}\:{username}\:“\pi.''{there} \\ $$$${It}\:{displayed}\:{question}\:{mark}\:{instead} \\ $$$${of}\:\pi\:{and}\:{showed}\:{the}\:{dot}\:{as}\:{it}\:{is} \\ $$
Commented by Physics lover last updated on 23/Oct/17
you can also repeat the experiment. you will get the same result. ;)
$${you}\:{can}\:{also}\:{repeat}\:{the}\:{experiment}. \\ $$$$\left.{you}\:{will}\:{get}\:\:{the}\:{same}\:{result}.\:;\right) \\ $$$$ \\ $$
Commented by Rasheed.Sindhi last updated on 23/Oct/17
Anyway you can write a monograph paper on IDology now. lol.
$$\mathrm{Anyway}\:\mathrm{you}\:\mathrm{can}\:\mathrm{write}\:\mathrm{a}\:\mathrm{monograph} \\ $$$$\mathrm{paper}\:\mathrm{on}\:\boldsymbol{\mathrm{ID}}\mathrm{ology}\:\mathrm{now}.\:{lol}. \\ $$
Commented by Rasheed.Sindhi last updated on 23/Oct/17
This will be a new subject introduced by you. hahaha.. lol.
$$\mathrm{This}\:\mathrm{will}\:\mathrm{be}\:\mathrm{a}\:\mathrm{new}\:\mathrm{subject}\:\mathrm{introduced} \\ $$$$\mathrm{by}\:\mathrm{you}.\:\mathrm{hahaha}..\:{lol}. \\ $$
Commented by Physics lover last updated on 23/Oct/17
lol
$$ \\ $$$${lol} \\ $$$$ \\ $$

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