The ionisation potential of hydrogen atom is 13.6 eV. The energy required to remove an electron in the n = 2 state of the helium atom is

Question

The ionisation potential of hydrogen atom is 13-6 eV- The energy required to remove an electrons in the n-2 state of hydrogen atom is-

Toppr Admin · Accepted Answer

Correct option is D- -3-4- eV-E-n-13-6- ev-left -dfrac -Z-2-n-2-right-dfrac -E-1-n-2-I-E-1 -E-1 -13-6- eV -Rightarrow E-1-13-6- eV-therefore E-2 -dfrac -13-6- ev-4-3-4ev-therefore I-E-2-E-2 -3-4- ev-Option D is correct-

The ionisation potential of hydrogen atom is $$13.6\ eV$$. The energy required to remove an electrons in the $$n=2$$ state of hydrogen atom is:

Correct option is D. $$3.4\ eV$$
$$E_n=-13.6\ ev\left (\dfrac {Z^2}{n^2}\right)=\dfrac {-E_1}{n^2}[I.E_1 =-E_1 =13.6\ eV \Rightarrow E_1=-13.6\ eV]$$

$$\therefore E_2 =\dfrac {-13.6\ ev}{4}=-3.4ev$$

$$\therefore I.E_2=-E_2 =3.4\ ev$$.

Option D is correct.

The ionisation potential of hydrogen atom is $$13.6\ eV$$. The energy required to remove an electrons in the $$n=2$$ state of hydrogen atom is:

Correct option is D. $$3.4\ eV$$$$E_n=-13.6\ ev\left (\dfrac {Z^2}{n^2}\right)=\dfrac {-E_1}{n^2}[I.E_1 =-E_1 =13.6\ eV \Rightarrow E_1=-13.6\ eV]$$$$\therefore E_2 =\dfrac {-13.6\ ev}{4}=-3.4ev$$$$\therefore I.E_2=-E_2 =3.4\ ev$$.Option D is correct.

Correct option is D. $$3.4\ eV$$
$$E_n=-13.6\ ev\left (\dfrac {Z^2}{n^2}\right)=\dfrac {-E_1}{n^2}[I.E_1 =-E_1 =13.6\ eV \Rightarrow E_1=-13.6\ eV]$$

$$\therefore E_2 =\dfrac {-13.6\ ev}{4}=-3.4ev$$

$$\therefore I.E_2=-E_2 =3.4\ ev$$.

Option D is correct.