JEE MAIN - Chemistry (2018 - 15th April Evening Slot - No. 21)
The de-Broglie's wavelength of electron present in first Bohr orbit of 'H' atom is :
0.529 $$\mathop {\rm A}\limits^ \circ $$
2$$\pi $$ $$ \times $$ 0.529 $$\mathop {\rm A}\limits^ \circ $$
$${{0.529} \over {2\pi }}$$ $$\mathop {\rm A}\limits^ \circ $$
4 $$ \times $$ 0.529 $$\mathop {\rm A}\limits^ \circ $$
Explanation
Radius r = 0.529 $$ \times $$ $${{{n^2}} \over z}$$
In first Bohr orbit of hydrogen atom radius,
r = 0.529 $$ \times $$ $${{{1^2}} \over 1}$$ = 0.529 $$\mathop A\limits^ \circ $$
Angular momentum,
mvr = $${{nh} \over {2\pi }}$$
$$\therefore\,\,\,$$ for n = 1,
mvr = $${h \over {2\pi }}$$
$$ \Rightarrow $$$$\,\,\,$$ 2$$\pi $$r = $${h \over {mv}}$$ = $$\lambda $$ [ as $$\lambda $$ = $${h \over {mv}}$$]
$$\therefore\,\,\,$$ $$\lambda $$ = 2$$\pi $$r = 2 $$\pi $$ $$ \times $$ 0.5 29
In first Bohr orbit of hydrogen atom radius,
r = 0.529 $$ \times $$ $${{{1^2}} \over 1}$$ = 0.529 $$\mathop A\limits^ \circ $$
Angular momentum,
mvr = $${{nh} \over {2\pi }}$$
$$\therefore\,\,\,$$ for n = 1,
mvr = $${h \over {2\pi }}$$
$$ \Rightarrow $$$$\,\,\,$$ 2$$\pi $$r = $${h \over {mv}}$$ = $$\lambda $$ [ as $$\lambda $$ = $${h \over {mv}}$$]
$$\therefore\,\,\,$$ $$\lambda $$ = 2$$\pi $$r = 2 $$\pi $$ $$ \times $$ 0.5 29
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