When photon of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV and de-Brolie wavelength $${{\lambda }_{A}}$$. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.70 eV is $${{T}_{B}}=({{T}_{A}}-1.50)\ eV$$. If the de-Broglie wavelength of these photoelectrons is $${{\lambda }_{B}}=2{{\lambda }_{A}}$$, then [IIT-JEE 1994]

Question

When photon of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV and de-Brolie wavelength $${{\lambda }_{A}}$$. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.70 eV is $${{T}_{B}}=({{T}_{A}}-1.50)\ eV$$. If the de-Broglie wavelength of these photoelectrons is $${{\lambda }_{B}}=2{{\lambda }_{A}}$$, then                                                              [IIT-JEE 1994]

TuteeHUB · Accepted Answer

$${{T}_{A}}=2.00\ eV$$

TuteeHUB · Answer

The work function of A is 2.25 eV

TuteeHUB · Answer

The work function of B is 4.20 eV

TuteeHUB · Answer

$${{T}_{B}}=2.75\ eV$$

1.	When photon of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV and de-Brolie wavelength \[{{\lambda }_{A}}\]. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.70 eV is \[{{T}_{B}}=({{T}_{A}}-1.50)\ eV\]. If the de-Broglie wavelength of these photoelectrons is \[{{\lambda }_{B}}=2{{\lambda }_{A}}\], then [IIT-JEE 1994]
A.	The work function of A is 2.25 eV
B.	The work function of B is 4.20 eV
C.	\[{{T}_{A}}=2.00\ eV\]
D.	\[{{T}_{B}}=2.75\ eV\]
Answer» C. \[{{T}_{A}}=2.00\ eV\]

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