Q: Light of frequency 1.5 times the threshold frequency is incident on photosensitive material. If the frequency is halved and intensity is doubled, the photo current becomes

When the frequency of incident light is halved of its original value i.e. \$1.5v_0\$ then it becomes less than the threshold value. In that case no photoelectric effect takes place. No photoelectrons would be emitted. The photoelectric current becomes zero.

Q: An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vacuum)

For electronDe - Broglie wavelength \$\lambda_c = \dfrac{h}{p}\$where p is momentum \$p = mv\$also by energy we have \$E = \dfrac{1}{2} mv^2\$\$\Rightarrow E = \dfrac{1}{2} \dfrac{p^2}{m}\$\$\Rightarrow p = \sqrt{2 mE}\$\$\therefore \lambda_c = \dfrac{h}{\sqrt{2 mE}}\$For photon energy \$\Rightarrow E = \dfrac{hc}{\lambda}\$\$\Rightarrow \lambda = \dfrac{h c}{E}\$\$\therefore \dfrac{\lambda_c}{\lambda} = \dfrac{h}{\sqrt{2 mE}} \dfrac{E}{hc}\$\$= \dfrac{1}{C} \sqrt{\dfrac{E}{2m}}\$option (A) is correct.

Q: Light of two different frequencies whose photons have energies 1eV and 2.5eV successively illuminate a metal of work function 0.5 eV.The ratio of the maximum speeds of the emitted electrons will he

For maximum speed of the photo electrons \$\dfrac{1}{2}mv^2 = E_{photon} - \phi\$where \$\phi = 0.5\$ eV is the work function of the metal.Energy of photon \$E_1 = 1eV\$\$\therefore\$ \$\dfrac{1}{2}mv_1^2 =1- 0.5\$We get \$\dfrac{1}{2}mv_1^2 =0.5\$ eV ....(1)Energy of photon \$E_2 = 2.5eV\$\$\therefore\$ \$\dfrac{1}{2}mv_2^2 =2.5- 0.5\$We get \$\dfrac{1}{2}mv_2^2 =2\$ eV ....(2)Ratio of maximum speeds \$\dfrac{v_1^2}{v_2^2} = \dfrac{0.5}{2} = \dfrac{1}{4}\$We get \$\dfrac{v_1}{v_2} = \dfrac{1}{2}\$

Q: A proton, a neutron, an electron and an \$\alpha\$-particle have same energy. Then their de Broglie wavelengths compare as :

Kinetic energy of particle, \$K = \dfrac{1}{2}mv^2 \,\,\,\, or \,\,\,\,\, mv = \sqrt{2mK}\$de Broglie wavelength, \$\lambda = \dfrac{h}{mv} = \dfrac{h}{\sqrt{2mK}}\$For the given value of \$K, \lambda \alpha \dfrac{1}{\sqrt{m}}\$\$\therefore \lambda_p : \lambda_n : \lambda_e : \lambda_{\alpha} = \dfrac{1}{\sqrt{m_p}} : \dfrac{1}{\sqrt{m_n}}: \dfrac{1}{\sqrt{m_e}} :\dfrac{1}{\sqrt{m_\alpha}}\$Since \$m_p = m_n, \,\, hence\,\, \lambda_p = \lambda_n\$\$ As\ m_{\alpha} > m_p, \,\, therefore \,\, \lambda_{alpha} < \lambda_p\$As \$m_e < m_n, therefore \lambda_n < \lambda_n\$Hence \$ \lambda_{alpha} < \lambda_p = \lambda_n < \lambda_e\$

Question 1

A proton and an alpha particle both are accelerated through the same potential difference. The ratio of corresponding de-Broglie wavelengths is :

Accepted Answer

Key Point : The de-Broglie wavelength of a particle of mass m and moving with velocity $v$ is given by$\lambda = \dfrac {h}{mv} $       $ (\because p = mv)$de-Broglie wavelength of a proton of mass $m_{1}$ and kinetic energy $k$ is given by$\lambda_{1} = \dfrac {h}{\sqrt {2m_{1}k}}$           $(\because p = \sqrt {2mk})$$\lambda_1= \dfrac {h}{\sqrt {2m_{1}qV}} .... (i)$        $ [\because k = qV]$For an alpha particle mass $m_{2}$ carrying charge $q_{0}$ is accelerated through potential $V$, then$\lambda_{2} = \dfrac {h}{\sqrt {2m_{2}q_{0}V}}$$\because$ For $\alpha - particle\  \   (^{4}_{2}He)$ :  $ q_{0} = 2q$ and $m_{2} = 4m_{1}$$\therefore \lambda_{2} = \dfrac {h}{\sqrt {2\times 4m_{1}\times 2q\times V}} .... (ii)$The ratio of corresponding wavelength, from Eqs. (i) and (ii), we get$\dfrac {\lambda_{1}}{\lambda_{2}} = \dfrac {h}{\sqrt {2m_{1}qV}} \times \dfrac {\sqrt {2\times m_{1} \times 4\times 2qV}}{h} = \dfrac {4}{\sqrt {2}} \times \dfrac {\sqrt {2}}{\sqrt {2}}$We get  $\dfrac{\lambda_1}{\lambda_2}= 2\sqrt {2}$

Question 2

Light of frequency 1.5 times the threshold frequency is incident on photosensitive material. If the frequency is halved and intensity is doubled, the photo current becomes

Accepted Answer

When the frequency of incident light is halved&#10;of its original value i.e. $1.5v_0$ then it becomes less than the threshold value. In&#10;that case no photoelectric effect takes place. No photoelectrons would be emitted.&#10;The photoelectric current becomes zero.

Question 3

An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vacuum)

Accepted Answer

For electronDe - Broglie wavelength $\lambda_c = \dfrac{h}{p}$where p is momentum $p = mv$also by energy we have $E = \dfrac{1}{2} mv^2$$\Rightarrow E = \dfrac{1}{2} \dfrac{p^2}{m}$$\Rightarrow p = \sqrt{2 mE}$$\therefore \lambda_c = \dfrac{h}{\sqrt{2 mE}}$For photon energy $\Rightarrow E = \dfrac{hc}{\lambda}$$\Rightarrow \lambda = \dfrac{h c}{E}$$\therefore \dfrac{\lambda_c}{\lambda} = \dfrac{h}{\sqrt{2 mE}} \dfrac{E}{hc}$$= \dfrac{1}{C} \sqrt{\dfrac{E}{2m}}$option (A) is correct.

Question 4

An electron, an alpha particle and a proton have the same kinetic energy. Which one of these particles has (i) the shortest and (ii) the largest, de Broglie wavelength?

Accepted Answer

De Broglie wavelength of particle of mass m in terms of kinetic energy $(E_k)$ is $\lambda=\dfrac{h}{\sqrt{2mE_k}} \propto \dfrac{1}{m}$ for the same kinetic energy.(i) Out of given particles, the mass of alpha particle is maximum, so de Broglie wavelength associated with alpha particle is the shortest.(ii) As mass of electron is least, so electron has largest de-Broglie wavelength.

Question 5

Draw graphs showing variation of photoelectric current with applied voltage for two incident radiations of equal frequency and different intensities. Mark the graph for the radiation of higher intensity.

Accepted Answer

The above graph is showing the variation of photoelectric current with the applied voltage. They have different intensities but same frequency.Here the upper one is for high intensity and lower one is for low intensity.

Question 6

Light of two different frequencies whose photons have energies 1eV and 2.5eV successively illuminate a metal of work function 0.5 eV.The ratio of the maximum speeds of the emitted electrons will he

Accepted Answer

For maximum speed of the photo electrons  $\dfrac{1}{2}mv^2  = E_{photon} - \phi$where $\phi = 0.5$ eV is the work function of the metal.Energy of photon $E_1 = 1eV$$\therefore$  $\dfrac{1}{2}mv_1^2  =1- 0.5$We get  $\dfrac{1}{2}mv_1^2  =0.5$ eV        ....(1)Energy of photon $E_2 = 2.5eV$$\therefore$  $\dfrac{1}{2}mv_2^2  =2.5- 0.5$We get  $\dfrac{1}{2}mv_2^2  =2$ eV        ....(2)Ratio of maximum speeds  $\dfrac{v_1^2}{v_2^2} = \dfrac{0.5}{2} = \dfrac{1}{4}$We get   $\dfrac{v_1}{v_2} =  \dfrac{1}{2}$

Question 7

Write four laws of photoelectric effect.

Accepted Answer

Photo electric effect: When some photosensitive substances are exposed to ultraviolet light, electrons are ejected from their surface which are called photoelectrons. This process is called photoelectric emission.Laws of photoelectric emission:(i) There is a definite cut off value of frequency below which electrons cannot be ejected by any substance.(ii) Number of emitted electrons are directly proportional to the intensity of light incident.(iii) Kinetic energy of emitted electrons depends on the frequency of incident light on substance.(iv) There is no time logging between the incident of light and emission of electrons.

Question 8

Which of the following phenomenon works on the particle nature of light?

Accepted Answer

(a) Interference is a phenomenon in which two waves of same frequency superpose to give resultant intensity different from sum of their separate intensity. So, it cannot exhibit particle's nature of light.(b) Diffraction is a phenomenon in which light bends at sharp ends of an obstacle or a hole. So, it also cannot exhibit particle's nature of light.(c) Polarisation of light is a property owing to which a light ray after emerging through a crystal (a special kind like tourmaline) have vibrations in a plane perpendicular to its direction of propagation. So, it also can not explain particle's nature of light.(d) Photoelectric effect states that light travels in the form of bundles or packets of energy, called photons. This effect is explained on the basis of quantum nature of light. So, it clearly explains the particles's nature of lighthence (d) is correct.

Question 9

What is photoelectric effect? Explain the effect of increase (i) frequency (ii) intensity of incident radiation on photoelectric current with suitable graphs.

Accepted Answer

Photoelectric effect is a phenomenon which is based on the idea that electromagnetic radiation is made of a series of particles called photons.When a photon hits an electron  on a metal surface,the electron can be emitted.The emitted electrons are called photoelectrons.(i)On increasing the frequency the saturation currents remains the same as seen in the first figure.(ii)On increasing the Intensity the saturation current also increases.

Question 10

A proton, a neutron, an electron and an $\alpha$-particle have same energy. Then their de Broglie wavelengths compare as :

Accepted Answer

Kinetic energy of particle, $K = \dfrac{1}{2}mv^2 \,\,\,\, or \,\,\,\,\, mv = \sqrt{2mK}$de Broglie wavelength, $\lambda = \dfrac{h}{mv} = \dfrac{h}{\sqrt{2mK}}$For the given value of $K, \lambda \alpha \dfrac{1}{\sqrt{m}}$$\therefore \lambda_p : \lambda_n : \lambda_e : \lambda_{\alpha} = \dfrac{1}{\sqrt{m_p}} : \dfrac{1}{\sqrt{m_n}}: \dfrac{1}{\sqrt{m_e}} :\dfrac{1}{\sqrt{m_\alpha}}$Since $m_p = m_n, \,\, hence\,\, \lambda_p = \lambda_n$$ As\ m_{\alpha} > m_p, \,\, therefore \,\, \lambda_{alpha} < \lambda_p$As $m_e < m_n, therefore \lambda_n < \lambda_n$Hence $ \lambda_{alpha} < \lambda_p = \lambda_n < \lambda_e$