NEET Practice Questions (MCQs) with Answers & Solutions

According to Rayleigh and Jeans, the black-body radiation in a cavity is a system of standing electromagnetic waves. This is because the radiation inside the cavity can be described as standing waves, which form due to the boundary conditions of the cavity.

$ {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 $