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

$ \phi = hf _0 $ $ = { hc \over \lambda_0 } J = { hc \over \lambda_0 e } eV $ $ \therefore \lambda_0 \alpha {1 \over 0 } $ $ \therefore {\varphi_0( tungsten) \over \varphi _0 (sodium) }= { \varphi (tungsten) \over \varphi (sodium) } ={ 2.3 \over 4.5} $ $ \therefore \lambda_0 tungsten = { 2.3 \over 4.5 } \times 5460 A ^ \circ $ $ = 2790.6 = 2791 A ^ \circ $

In the Davisson and Germer experiment, the filament of the electron gun is coated with barium oxide (BaO). This coating helps in the efficient emission of electrons when heated.

The probability of finding a particle at a given point in space is directly proportional to the square of the wave function's magnitude, $| \\Psi|^2$. This is because the probability density is given by $\\rho(x,y,z) = | \\Psi(x,y,z)|^2$. Therefore, the correct option is 'directly proportional to $ | \\Psi \\Psi ^ * | $'.

$ E = hf = h { C \over \lambda } $ $ = { 6.62 \times 10^ {-34} \times 3.0 \times 10^8 } $ $ = 3.31 \times 10^ {-19 } J $ $ = { 3.31 \times 10^{-19 } \over 1.6 \times 10 ^ {-19 }} eV $ $ = 2.06 eV $

$ hf_0 = \phi $ $ \therefore f_0 = { \phi \over h } = { 5.3 \times 10^{-19} \times 1.6 \over 6.62 \times 10^ {-34} } Hz $ $ = 1.3 \times 10^{15 } Hz $

$ eV_0 = { 1 \over 2 } mv^2 max = 3.8 eV $ $ \therefore V_0 = 3.8 V $ $ {1 \over 2 } mv^2 max = eV_0 = hf - \phi = hf - hf_0 = h ({c \over \lambda} - {c \over \lambda_0} ) $

When the frequency of the incident radiation is increased, while keeping the intensity constant, the energy of each photon increases according to the relation $E = h\nu$ (where $\nu$ is the frequency and $h$ is Planck's constant). Therefore, the energy of the emitted photoelectrons will increase.