Question

estimate the difference in energy between the 1st and 2nd bohr orbit for hydrogen atom.at what minimum atomic number would a transition from n=2 to n=1 energy level result in the emission of X-rays with ^=3.0*10^-8m?which hydrogen like species does thi atomic number corresponds to?

Answer 1

@Frostbite

Answer 2

@cambrige

Answer 3

@Kayne @.Sam. @TheDoctorRules

Answer 4

thnx buddy

Answer 5

\[1/λ=1.09667∗10^{−7}(Z)(1/n _{i}^2−1/n _{f}^2)\]

Answer 6

U got the answer from that??O_0

Answer 7

doing it W8..

Answer 8

Substitue your lamdba value and ni and nf values find the value of Z

Answer 9

Wat do u get Z as?? Take ur time!1

Answer 10

This equation was for the 2nd part of the problem!!

Answer 11

For the first part use

Answer 12

\[\Delta E={R _{H}}({1/n1^2-1/n2^2})\]

Answer 13

i got that Z=0.101

Answer 14

Where \[R _{H}=-2.18*10^{-18} J\]

Answer 15

Sure??

Answer 16

Is 3*10^-8 the wavelength or frequency??

Answer 17

it,s wavelength

Answer 18

yeah sure

Answer 19

I did a mistake...Its 1.09*10^7 not 10^-7

Answer 20

Sorry

Answer 21

Okk then i got Z=1.01*10^-2

Answer 22

U get Z=4 so It is Helium!!

Answer 23

U dint multiply 1.01 with 4!!

Answer 24

And how did u get 10^-2???O_O

Answer 25

\[10^8/3=1.09667*10^7*Z*(1/n _{i}^2-1/n _{f}^2)\]

Answer 26

Solve this and get me the ans!!

Answer 27

Its square of n(i) and n(f) giving u 4 in the denominator!!lol

Answer 28

Using xact vlues u get approx 4.0528 i.e 4! So this atomic number belongs to helium!!
Get me ??

Answer 29

@hamza10 u there??

Answer 30

@cambrige sorry i was a little busy

Answer 31

yes i got u