JEE CHEMISTRY MCQ: STRUCTURE OF ATOM

1 Hard

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Analyzing +4 | -1

When photon of energy

4.25 \mathrm{~eV}

strikes the surface, the ejected electron has maximum kinetic energy

T_A

expressed in

\mathrm{eV}

and de Broglie wavelength

\lambda_A.

The maximum kinetic energy of photoelectrons liberated by another metal

B

by photons of

4.70 \mathrm{~eV}

T_B

\left(T_B=T_A-1.5 \mathrm{~eV}\right).

If de Broglie wavelength of the electron is

\lambda_B\left(\lambda_B=2 \lambda_A\right),

then which is not correct?

View answer and explanation

Correct option: C

Work function of

B

1.20 \mathrm{~eV}

Option D (A is correct but R is not correct) is correct because the basicity decreases due to the lanthanide contraction, but the reason given about covalent character is incorrect.

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