Dual Nature of Matter and Radiation MCQs for NEET — Physics Questions with Answers

In the context of electromagnetic radiation in a cavity, the energy of an oscillator being equal to $5hf$ indicates that the oscillator has energy levels that can be occupied by photons. Since the energy is an integer multiple of $hf$, it suggests that the oscillator has a one-to-one correspondence with the photons. Thus, the oscillator is equivalent to 1:1.

In metals, valence electrons can move freely within the metal but they are not completely free. They move according to their wave functions inside the metal. This is due to the quantum mechanical nature of electrons which describes their behavior in terms of wave functions.

$ {4 \pi \times 10^ {20 } }$ $ \triangle x = 2 r = 2 \times 10^{-15} m$ $ \triangle x. \triangle p \approx { h \over 2 \pi } $ $ = { 66 \times 10 ^ {-34} \over 2 \times 2 \times 3.14 \times 2 \times 10h -15 } = 0.5255 \times 10^{-19} $ $ E ={ p^2 \over 2m } P \approx \triangle p $ $ = { (0.5255 \times 10^{-19} )^2 \over 2 \times 9.1 \times 10^{-31} }J = { (0.5255 \times 10^{-19} )^2 \over 2 \times 9.1 \times 10^ {-31} \times 1.6 \times 10^{-19}} $ $ E = 9.48 \times 10^ 3 MeV $

$3.56 \mu A$ $ P = { E \over t} $ $ P = {n_1 hc \over \lambda } . t $ $ \therefore n_1 = { p \lambda \over hct } = { 2 \times 10^{-9} \times 44 \times 10 ^ {-8} \over 6.6 \times 10^{-34} \times 3 \times 10^{8} \times 1 } = 4.44 \times 10^9 $ $ n = n_1 of 0.5 \% = 4.4 \times 10^9 \times {0.5 \over 100} $ $ I = ne = 2.22 \times 10h7 \times 1.6 \times 10h -19 = 3.552 $ $ I = 3.56 \times 10^{-6} \mu A$

$ E_{K1} - E _{K2} = hc ( {1 \over \lambda_1} - { 1 \over \lambda_2 } ) = { hc (\lambda_2 -\lambda_1) \over \lambda_1 \lambda_2 }$ $ \therefore E_{K1} - E_{K2} $ $ = { (6.62 \times 10^{-34}) \times (3 \times 10^8 ) ( 5000 -2500 ) \times 10^{-10} \over (5000 \times 10^{-10} \times (25000) \times 10^{-10}}$ $ = 3.96 \times 10^{-19} J $

Photoelectric emission occurs only when the incident light has more than a certain minimum frequency. This is because only photons with energy greater than the work function of the material can cause the emission of electrons. The energy of a photon is given by the formula $E = hf$, where $h$ is Planck's constant and $f$ is the frequency. Thus, a minimum frequency is required to provide sufficient energy to overcome the work function and release electrons.

$ \phi _0 = hf $ $ \therefore f_0 = { \phi_0 \over h} ={ 3.3 \times 1.6 \times 10^{-19} \over 6.6 \times 10^{-34} }$ $ \therefore f_0 = 0.8 \times 10^{15} $ $ \therefore f_0 = 8.0 \times 10^{14} Hz $