JEE MAIN - Physics (2019 - 12th April Evening Slot - No. 16)
Consider an electron in a hydrogen atom revolving in its second excited state (having radius 4.65 $$\mathop A\limits^o $$). The
de-Broglie wavelength of this electron is :
6.6 $$\mathop A\limits^o $$
3.5 $$\mathop A\limits^o $$
9.7 $$\mathop A\limits^o $$
12.9 $$\mathop A\limits^o $$
Explanation
For second excited state n = 3
$$mvr = {{3h} \over {2\pi }}$$ ........(1)
$$mv = {h \over \lambda }$$ .........(2)
Dividing (1) by (2), we get
$$r = {{3\lambda } \over {2\pi }}$$
$$ \Rightarrow $$ $$\lambda = {{2\pi r} \over 3}$$ = $${{2 \times 3.14 \times 4.65} \over 3}$$ = 9.7 $$\mathop A\limits^o $$
$$mvr = {{3h} \over {2\pi }}$$ ........(1)
$$mv = {h \over \lambda }$$ .........(2)
Dividing (1) by (2), we get
$$r = {{3\lambda } \over {2\pi }}$$
$$ \Rightarrow $$ $$\lambda = {{2\pi r} \over 3}$$ = $${{2 \times 3.14 \times 4.65} \over 3}$$ = 9.7 $$\mathop A\limits^o $$
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