Question

$\bf{\mathfrak{The \: wavelength \: of \: radiation \: emitted \: when \: in \: {He}^{+} \: electron \: falls \: from \: infinity \: to \: stationary \: state \: would \: be \: (R = 1.097 \times {10}^{7} \: {m}^{-1}\ \textless \ br /\ \textgreater \ }}$

$\bf{\mathfrak{(1) \: 2.2 \times {10}^{-8} \: m}}$
$\bf{\mathfrak{(2) \: 2.2 \times {10}^{-9} \: m}}$
$\bf{\mathfrak{(3) \: 120 \: m}}$
$\bf{\mathfrak{(4) \: 22 \times {10}^{7} \: m}}$

Answer 1

We know that,

$\frac{1}{\lambda}= Rz^{2}(\frac{1}{(n_1)^{2}} - \frac{1}{(n_1)^{2}}) \\ \\ \frac{1}{\lambda}= R2^{2}(\frac{1}{1^{2}} - \frac{1}{\infty}) \\ \\ \frac{1}{\lambda}= 4 \times 1.097 \times 10^{7} (1 - 0) \\ \\ \frac{1}{\lambda}= 4.388 \times 10^{7} \\ \\ \lambda = \frac{1}{ 4.388 \times 10^{7}} \\ \\ \lambda = 0.227 \times 10^{-7} \\ \\ \lambda = 2.2 \times 10^{-8} m$

Answer 2

HEY....

HERE'S YOUR....

HIGH RATED GABRU---HARSH ❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️

No , it is not necessary that initially it will be at infinity level . It may be present at any level say n= 2,3,4,5,etc . Now as radiation is emitted , transition must be from higher level to any lower level which can determined by knowing the wavelength of emitted radiation . If you measure the wavelength of emitted radiation by any instrument , then after knowing it's value , you can determine the transition of electron by following formula ,

1/wavelength =109677[1/n^2 - 1/m^2]

Here , unit of wavelength is cm^-1 .

n is initial state of electron and m is final state of electron .

n,m are integers only .

❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️