JEE MAIN - Physics (2021 - 24th February Evening Shift - No. 19)
According to Bohr atomic model, in which of the following transitions will the frequency be maximum?
n = 2 to n = 1
n = 3 to n = 2
n = 4 to n = 3
n = 5 to n = 4
Explanation
Let, nf, ni be the final and initial orbit.
As we know that,
$${1 \over \lambda } = 1.09 \times {10^7}\left[ {{1 \over {n_f^2}} - {1 \over {n_i^2}}} \right]$$
Now, checking for each option, we get
(a) $${1 \over \lambda } \propto \left[ {{1 \over {{3^2}}} - {1 \over {{4^2}}}} \right] = \left[ {{1 \over 9} - {1 \over {16}}} \right] = 0.05$$ .... (i)
(b) $${1 \over \lambda } \propto \left[ {{1 \over 1} - {1 \over 4}} \right] = 0.75$$ .... (ii)
(c) $${1 \over \lambda } \propto \left[ {{1 \over {16}} - {1 \over {25}}} \right] = 0.0225$$ .... (iii)
(d) $${1 \over \lambda } \propto \left[ {{1 \over 4} - {1 \over 9}} \right] = 0.14$$ .... (iv)
The option (b) has highest value.
Since, frequency, $$f = {c \over \lambda } \Rightarrow f \propto {1 \over \lambda }$$
$$\therefore$$ Frequency will be maximum for transition n = 2 to n = 1.
As we know that,
$${1 \over \lambda } = 1.09 \times {10^7}\left[ {{1 \over {n_f^2}} - {1 \over {n_i^2}}} \right]$$
Now, checking for each option, we get
(a) $${1 \over \lambda } \propto \left[ {{1 \over {{3^2}}} - {1 \over {{4^2}}}} \right] = \left[ {{1 \over 9} - {1 \over {16}}} \right] = 0.05$$ .... (i)
(b) $${1 \over \lambda } \propto \left[ {{1 \over 1} - {1 \over 4}} \right] = 0.75$$ .... (ii)
(c) $${1 \over \lambda } \propto \left[ {{1 \over {16}} - {1 \over {25}}} \right] = 0.0225$$ .... (iii)
(d) $${1 \over \lambda } \propto \left[ {{1 \over 4} - {1 \over 9}} \right] = 0.14$$ .... (iv)
The option (b) has highest value.
Since, frequency, $$f = {c \over \lambda } \Rightarrow f \propto {1 \over \lambda }$$
$$\therefore$$ Frequency will be maximum for transition n = 2 to n = 1.
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