{"kind":"task","effective_mode":"full","benchmark":{"kind":"benchmark","effective_mode":"full","slug":"mmlu-pro","formal_name":"MMLU-Pro","introduction":"MMLU-Pro rebuilds MMLU with ten answer options instead of four and removes items that no longer separate models. Questions span fourteen academic and professional subjects.","introduction_ja":"","introduction_en":"","category":"Category not supplied","task_count":null,"acquisition_status":"Acquisition status not supplied","official_url":"https://huggingface.co/datasets/TIGER-Lab/MMLU-Pro","indexing_mode":"index","profile":{"resources":[],"task_format":"","scoring":"","metric":"","size":"","answer_access":"","license":"","citation":"","maintainer":"","released":"","why_hard":"","related":[],"history":"","construction":"","limitations":"","saturation":"","scores":[]}},"task_id":"a978f74f-1d8d-50fc-99c1-e4977adf18e1","task_key":"default--test--4331","task_revision_id":"1","upstream_id":"4331","short_description":"You are given that the ionization energy of a hydrogen atom is 13.527eV,…","config":"default","split":"test","body":"{\"category\":\"chemistry\",\"options\":\"['13.527eV', '162.324eV', '21.654eV', '108.216eV', '40.581eV', '54.108eV', '30.081eV', '72.144eV', '96.192eV', '27.054eV']\",\"question\":\"You are given that the ionization energy of a hydrogen atom is 13.527eV, calculate the second ionization energy of a helium atom. The general equation for the lines in the emis-sion spectra of one electron system such as H, He^+ , Li^2+ and Be^3+ is given as ѵ= Z^2 R_H{(1/n^2_1) - (1/n^2_2)} whereѵ= the wave numbers of emission lines and Z is the nuclear charge of the atom concerned.\",\"src\":\"stemez-PhysicalChemistry\"}","display_format":"text","language":"","answer_status":"published","assets":[],"source_url":"https://huggingface.co/datasets/TIGER-Lab/MMLU-Pro","history":"initial import","indexing_mode":"index","subproblems":[],"grids":[]}