JEE Main 2017 — Structure Of Atom Question with Solution

Note :

(1)   In Lyman Series, transition happens in n = 1 state
from n = 2, 3, . . . . . $$\propto$$

(2)   In Balmer Series, transition happens in n = 2 state
from n = 3, 4, . . . . . $$\propto$$

(3)   In Paschen Series, transition happens in n = 3 state
from n = 4, 5, . . . . . $$\propto$$

(4)   In Bracktt Series, transition happens in n = 4 state
from n = 5, 6 . . . . . . $$\propto$$

(5)   In Pfund Series, transition happens in n = 5 state
from n = 6, 7, . . . . $$\propto$$

We know,

     $$\frac{1}{\lambda }$$ = Rz² ($$\frac{1}{n_1^2}$$ $$-$$ $$\frac{1}{n_2^2}$$)

The shortest wavelength of hydrogen atom in Lyman series is from n₁ = 1 to n₂ = $$\propto$$

$$\therefore$$ $$\frac{1}{A}$$ = 1² R ($$\frac{1}{1^2}$$ $$-$$ $$\frac{1}{{\propto }^2}$$)    [for hydrogen, z = 1]

$$\Rightarrow$$ $$\frac{1}{A}$$ = R

The longest wavelength in pascal series of He⁺ is from n₁ = 3 to n₂ = 4

For He⁺, z = 2.

$$\frac{1}{\lambda }$$ = RZ² ($$\frac{1}{n_1^2}$$ $$-$$ $$\frac{1}{n_2^2}$$)

$$\Rightarrow$$ $$\frac{1}{\lambda }$$ = $$\frac{1}{A}$$ (2)² ($$\frac{1}{3^2}$$ $$-$$ $$\frac{1}{4^2}$$)

$$\Rightarrow$$ $$\frac{1}{\lambda }$$ = $$\frac{4}{A}$$ $$\left(\frac{7}{16\times 9}\right)$$ = $$\frac{7}{36A}$$

$$\Rightarrow$$ $$\lambda$$ = $$\frac{36A}{7}$$