SATHEE: Structure of Atom - Result Question 20

Structure of Atom - Result Question 20

####20. If ionization potential for hydrogen atom is $13.6 e V$ , then ionization potential for $H e^{+}$ will be

(a) $54.4 e V$

(b) $6.8 e V$

(d) $24.5 e V$

[1993]

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Solution:

(a) The ionization energy of any hydrogen like species (having one electron only) is given by the equation

I.E $= \frac{2 π^{2} Z^{2} m e^{4}}{h^{2}}$ or I.E $\propto Z^{2}$

Since the atomic number of $H$ is 1 and that of $H e$ is 2 , therefore, the I.E. of $H e^{+}$ is four times $(2^{2})$ the I.E. of Hi.e., $13.6 \times 4 = 54.4 e V$

For H-like particles,

$E_{n} = - \frac{21.8 \times 10^{- 19}}{h^{2}} Z^{2} J /$ atom

$= - \frac{1312}{n^{2}} Z^{2} k J / m o l$

$I . E = E_{a_{o}} - E_{1} = 0 - (- I \cdot E_{H} Z^{2})$

$= Z^{2} \times I_{E} H$

Now for $H e^{+}, Z = 2 ∴ I . E = 4 \times I_{H}$

$for L i^{2 +}, Z = 3 ∴ I . E = 9 \times I_{H}$

Structure of Atom - Result Question 20

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Structure of Atom - Result Question 20

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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