Atomic Structure Question 257
Question: The ionization energy of hydrogen atom is $ -13.6,eV. $ The energy required to excite the electron in a hydrogen atom from the ground state to the first excited state is (Avogadro-s constant = 6.022 × 1023)
Options:
A) $ 1.69\times {10^{-20}}J $
B) $ 1.69\times {10^{-23}}J $
C) $ 1.69\times 10^{23}J $
D) $ 1.69\times 10^{25}J $
Show Answer
Answer:
Correct Answer: B
Solution:
$ E=\frac{-13.6}{n^{2}}=\frac{-13.6}{4}=-3.4,eV $ We know that energy required for excitation $ \Delta E=E_2-E_1 $
$ =-3.4-(-13.6)=10.2,eV $ Therefore energy required for excitation of electron per atom $ =\frac{10.2}{6.02\times 10^{23}}=1.69\times {10^{-23}}J $
BETA
