JEE MAIN - Chemistry (2019 - 12th January Evening Slot - No. 15)
If the de Broglie wavelength of the electron in nth Bohr orbit in a hydrogenic atom is equal to 1.5 $$\pi $$a0 (a0 is Bohr radius), then the value of n/z is -
0.75
0.40
1.50
1.0
Explanation
According to debroglie hypothesis,
2$$\pi $$rn = n$$\lambda $$
$$ \therefore $$ $$\lambda $$ = $${{2\pi {r_n}} \over n}$$
According to the question,
$${{2\pi {r_n}} \over n}$$ = 1.5 $$\pi $$a0
We know,
rn = 0.529 $${{{n^2}} \over Z}$$
= a0 $$ \times $$ $${{{n^2}} \over Z}$$
$$ \therefore $$ $${{2\pi } \over n} \times {{{a_0}{n^2}} \over Z}$$ = 1.5 $$\pi $$ a0
$$ \Rightarrow $$ $${n \over Z}$$ = 0.75
2$$\pi $$rn = n$$\lambda $$
$$ \therefore $$ $$\lambda $$ = $${{2\pi {r_n}} \over n}$$
According to the question,
$${{2\pi {r_n}} \over n}$$ = 1.5 $$\pi $$a0
We know,
rn = 0.529 $${{{n^2}} \over Z}$$
= a0 $$ \times $$ $${{{n^2}} \over Z}$$
$$ \therefore $$ $${{2\pi } \over n} \times {{{a_0}{n^2}} \over Z}$$ = 1.5 $$\pi $$ a0
$$ \Rightarrow $$ $${n \over Z}$$ = 0.75
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