When photons of wavelength λ1 are incident on an isolated sphere, the corresponding stopping potential is found to be V.  When photons of wavelength  λ2 are used, the corresponding stopping potential was thrice that of the above value. If light of wavelength  λ3 is used then find the stopping potential for this case :

Option 1)

\frac{hc}{e}\left [ \frac{1}{\lambda _{3}} - \frac{1}{\lambda _{2}} - \frac{1}{\lambda _{1}}\right ]

Option 2)

\frac{hc}{e}\left [ \frac{1}{\lambda _{3}} + \frac{1}{\lambda _{2}} - \frac{1}{\lambda _{1}}\right ]

Option 3)

\frac{hc}{e}\left [ \frac{1}{\lambda _{3}} + \frac{1}{2\lambda _{2}} - \frac{3}{2\lambda _{1}}\right ]

Option 4)

\frac{hc}{e}\left [ \frac{1}{\lambda _{3}} + \frac{1}{2\lambda _{2}} - \frac{1}{\lambda _{1}}\right ]

Answers (2)
P perimeter

As we learnt in

Conservation of energy -

h\nu = \phi _{0}+\frac{1}{2}mv^{2}_{max}

h\nu = h\nu _{0}+\frac{1}{2}mv^{2}_{max}

 

h\left ( \nu-\nu _{0} \right )=\frac{1}{2}mv^{2}_{max}

where, h - Plank's constant\ \nu - Frequency\ \nu_{0} - threshold\ frequency\ \phi_{0}- work function

- wherein

 

 Applying Einstein photo electric equation 

\frac{1}{2}mv^{2}=\frac{hc}{\lambda_{1}}-{\phi}=eV                         .............(1)

\frac{hc}{\lambda_{2}}-{\phi}=3eV                                        .................(2)

Put wave of eV from equation (1) to equation (2)

\frac{hc}{\lambda_{2}}-{\phi}=3.\frac{hc}{\lambda_{1}}-3{\phi}

\Rightarrow\ \; {\phi}=\frac{1}{2}\left[\frac{3h}{\lambda_{1}}- \frac{hc}{\lambda_{2}} \right ]

If wavelength {\lambda_{3}} is used equation become 

eV'=\frac{hc}{\lambda_{3}}-{\phi}=\frac{hc}{\lambda_{3}}-\frac{1}{2}.\frac{3hc}{\lambda_{1}}+\frac{hc}{2{\lambda_{2}}}

V'=\frac{hc}{e}{\left[\frac{1}{\lambda_{3}}-\frac{3}{2\lambda_{1}}+\frac{1}{2\lambda_{2}} \right ]}

Correct option is option 3.


Option 1)

\frac{hc}{e}\left [ \frac{1}{\lambda _{3}} - \frac{1}{\lambda _{2}} - \frac{1}{\lambda _{1}}\right ]

This is incorrect option.

Option 2)

\frac{hc}{e}\left [ \frac{1}{\lambda _{3}} + \frac{1}{\lambda _{2}} - \frac{1}{\lambda _{1}}\right ]

This is incorrect option.

Option 3)

\frac{hc}{e}\left [ \frac{1}{\lambda _{3}} + \frac{1}{2\lambda _{2}} - \frac{3}{2\lambda _{1}}\right ]

This is the correct option.

Option 4)

\frac{hc}{e}\left [ \frac{1}{\lambda _{3}} + \frac{1}{2\lambda _{2}} - \frac{1}{\lambda _{1}}\right ]

This is incorrect option.

P perimeter

As we learnt in

Conservation of energy -

h\nu = \phi _{0}+\frac{1}{2}mv^{2}_{max}

h\nu = h\nu _{0}+\frac{1}{2}mv^{2}_{max}

 

h\left ( \nu-\nu _{0} \right )=\frac{1}{2}mv^{2}_{max}

where, h - Plank's constant\ \nu - Frequency\ \nu_{0} - threshold\ frequency\ \phi_{0}- work function

- wherein

 

 Applying Einstein photo electric equation 

\frac{1}{2}mv^{2}=\frac{hc}{\lambda_{1}}-{\phi}=eV                         .............(1)

\frac{hc}{\lambda_{2}}-{\phi}=3eV                                        .................(2)

Put wave of eV from equation (1) to equation (2)

\frac{hc}{\lambda_{2}}-{\phi}=3.\frac{hc}{\lambda_{1}}-3{\phi}

\Rightarrow\ \; {\phi}=\frac{1}{2}\left[\frac{3h}{\lambda_{1}}- \frac{hc}{\lambda_{2}} \right ]

If wavelength {\lambda_{3}} is used equation become 

eV'=\frac{hc}{\lambda_{3}}-{\phi}=\frac{hc}{\lambda_{3}}-\frac{1}{2}.\frac{3hc}{\lambda_{1}}+\frac{hc}{2{\lambda_{2}}}

V'=\frac{hc}{e}{\left[\frac{1}{\lambda_{3}}-\frac{3}{2\lambda_{1}}+\frac{1}{2\lambda_{2}} \right ]}

Correct option is 3.


Option 1)

\frac{hc}{e}\left [ \frac{1}{\lambda _{3}} - \frac{1}{\lambda _{2}} - \frac{1}{\lambda _{1}}\right ]

This is incorrect option.

Option 2)

\frac{hc}{e}\left [ \frac{1}{\lambda _{3}} + \frac{1}{\lambda _{2}} - \frac{1}{\lambda _{1}}\right ]

This is incorrect option.

Option 3)

\frac{hc}{e}\left [ \frac{1}{\lambda _{3}} + \frac{1}{2\lambda _{2}} - \frac{3}{2\lambda _{1}}\right ]

This is the correct option.

Option 4)

\frac{hc}{e}\left [ \frac{1}{\lambda _{3}} + \frac{1}{2\lambda _{2}} - \frac{1}{\lambda _{1}}\right ]

This is incorrect option.

Exams
Articles
Questions